Chapter 1 Sequences and Series - BS Publications1.3 Tests for the Convergence of an Infinite Series In order to study the nature of any given infinite series of +ve terms regarding

• Sequences and Series • Convergence of Infinite Series • Tests of Convergence

• P-Series Test • Comparison Tests • Ratio test • Raabe’s test • Cauchy’s Root test • Integral test • Leibnitz’s test

• Absolute Convergence • Conditional convergence • Power series and Interval of convergence

• Summary of all Tests • Solved University Questions (JNTU) • Objective type of Questions

Chapter 1 Sequences and Series

Engineering Mathematics - I

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1.1 Sequence A function f:N → S, where S is any nonempty set is called a Sequence i.e., for each n∈ N, ∃ a unique element f(n) ∈ S. The sequence is written as f(1), f(2), f(3), ......f(n)...., and is denoted by {f(n)}, or <f(n)>, or (f(n)). If f(n) = na , the sequence is written as 1 2, ..... na a a and denoted by , { } ( ).< >n n na or a or a Here f(n) or na are the

thn terms of the Sequence.

Ex. 1. 1 , 4 , 9 , 16 ,......... 2n ,.....(or) 2n< >

1

23. .

.

.

.

.

.

.n

49

2n

N S

Ex. 2. 3 3 3 3 3

1 1 1 1 1, , ,..... ....( )1 2 3

⎛ ⎞⎜ ⎟⎝ ⎠

orn n

Ex. 3. 1, 1, 1......1..... or <1>

Sequences and Series

3

Ex 4: 1 , –1, 1, –1, ......... or ( ) 11 −− n

Note : 1. If S ⊆ R then the sequence is called a real sequence. 2. The range of a sequence is almost a countable set.

1.1.1 Kinds of Sequences

1. Finite Sequence: A sequence na< > in which 0 na n m N= ∀ > ∈ is said to be a finite Sequence. i.e., A finite Sequence has a finite number of terms.

2. Infinite Sequence: A sequence, which is not finite, is an infinite sequence.

1.1.2 Bounds of a Sequence and Bounded Sequence

1. If ∃ a number ‘M’ na∋ ≤ M, n∀ ∈ N, the Sequence na< > is said to be bounded above or bounded on the right.

Ex. 1 11, , ,2 3

,....... here 1 na n≤ ∀ ∈N

2. If ∃ a number ‘m’ ,na m n∋ ≥ ∀ ∈N, the sequence na< > is said to be bounded below or bounded on the left.

Ex. 1 , 2 , 3 ,.....here 1 na n≥ ∀ ∈N 3. A sequence which is bounded above and below is said to be bounded.

Ex. Let ( ) 11 1⎛ ⎞= − +⎜ ⎟⎝ ⎠

nna

n

n 1 2 3 4 ......

na -2 3/2 -4/3 5/4 ......


4

From the above figure (see also table) it can be seen that m = –2 and M = 3

2.

∴ The sequence is bounded.

1.1.3 Limits of a Sequence

A Sequence na< > is said to tend to limit ‘l’ when, given any + ve number ' ',∈ however small, we can always find an integer ‘m’ such that ,− <∈ ∀ ≥na l n m , and we write nn

Lt a l→∞

= or →na l

Ex. If 2

2

12 3

+=

+nnan

then 12

< >→na .

1.1.4 Convergent, Divergent and Oscillatory Sequences 1. Convergent Sequence: A sequence which tends to a finite limit, say ‘l’ is

called a Convergent Sequence. We say that the sequence converges to ‘l’ 2. Divergent Sequence: A sequence which tends to ±∞ is said to be Divergent

(or is said to diverge). 3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is

called an Oscillatory Sequence.

Ex. 1. Consider the sequence 2 , 3 4 5, , ,.....2 3 4

here 11= +nan

The sequence < >na is convergent and has the limit 1

1 11 1 1− = + − =nan n

and 1<∈

n whenever 1

>∈

n

Suppose we choose .001∈= , we have 1 .001<n

when n > 1000.

Ex. 2. If ( ) 13 1' '

= + − < >nn na a

n converges to 3.


5

Ex. 3. If ( )2 1 . ,= + − < >nn na n n a diverges.

Ex. 4. If ( )1 2 1 ,= + − < >nn na a

n oscillates between -2 and 2.

1.2 Infinite Series If nu< > is a sequence, then the expression 1 2 3 ........ .....nu u u u+ + + + + is called an

infinite series. It is denoted by 1

∞

=Σ nn

u or simply Σ nu

The sum of the first n terms of the series is denoted by ns i.e., 1 2 3 1 2 3...... ; , , ,....= + + + +n n ns u u u u s s s s are called partial sums.

1.2.1 Convergent, Divergent and Oscillatory Series

Let Σ nu be an infinite series. As ,→ ∞n there are three possibilities. (a) Convergent series: As ,→ ∞ →nn s a finite limit, say ‘s’ in which case the

series is said to be convergent and ‘s’ is called its sum to infinity.

Thus →∞

=nnLt s s (or) simply nLts s=

This is also written as 1 2 3 ..... ... .nu u u u to s+ + + + + ∞ = (or) 1

∞

=Σ =nn

u s (or)

simply .Σ =nu s (b) Divergent series: If → ∞ns or −∞ , the series said to be divergent. (c) Oscillatory Series: If ns does not tend to a unique limit either finite or infinite it

is said to be an Oscillatory Series.

Note: Divergent or Oscillatory series are sometimes called non convergent series.

1.2.2 Geometric Series

The series, 2 11 ..... ...−+ + + +nx x x is

(i) Convergent when 1<x , and its sum is 11− x

(ii) Divergent when 1≥x .

(iii) Oscillates finitely when x = -1 and oscillates infinitely when x < -1.

Proof: The given series is a geometric series with common ratio ‘x’

∴ 11−

=−

n

nxsx

when 1≠x [By actual division – verify]


6

(i) When 1:<x

1 11 1 1→∞ →∞ →∞

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟− − −⎝ ⎠ ⎝ ⎠

n

nn n n

xLt s Lt Ltx x x

since 0 as ⎡ ⎤→ → ∞⎣ ⎦nx n

∴ The series converges to 11− x

(ii) When 11:1

−≥ =

−

n

nxx sx

and → ∞ns as → ∞n

∴ The series is divergent. (iii) When x = –1: when n is even, 0→ns and when n is odd, 1→ns ∴ The series oscillates finitely. (iv) When 1,< − → ∞nx s or −∞ according as n is odd or even. ∴ The series oscillates infinitely.

1.2.3 Some Elementary Properties of Infinite Series 1. The convergence or divergence of an infinites series is unaltered by an addition or

deletion of a finite number of terms from it. 2. If some or all the terms of a convergent series of positive terms change their signs,

the series will still be convergent. 3. Let Σ nu converge to ‘s’ Let ‘k’ be a non – zero fixed number. Then Σ nku converges to ks. Also, if Σ nu diverges or oscillates, so does Σ nku

4. Let Σ nu converge to ‘l’ and Σ nv converge to ‘m’. Then (i) ( )Σ +n nu v converges to ( l + m ) and (ii) ( )Σ +n nu v converges to ( l – m )

1.2.4 Series of Positive Terms Consider the series in which all terms beginning from a particular term are +ve.

Let the first term from which all terms are +ve be 1u . Let Σ nu be such a convergent series of +ve terms. Then, we observe that the convergence is unaltered by any rearrangement of the terms of the series.

1.2.5 Theorem

If Σ nu is convergent, then 0→∞

=nnLt u .

Proof : 1 2 ......= + + +n ns u u u 1 1 2 1,......− −= + + +n ns u u u , so that, 1−= −n n nu s s


7

Suppose Σ =nu l then →∞

=nnLt s l and 1−→∞

=nnLt s l

∴ ( )1−→∞ →∞= −n n nn n

Lt u Lt s s ; 1 0−→∞ →∞− = − =n nn n

Lt s Lt s l l

Note: The converse of the above theorem need not be always true. This can be observed from the following examples.

(i) Consider the series, 1 1 11 ....... ....2 3

+ + + + +n

; 1 , 0→∞

= =n nnu Lt u

n

But from p-series test (1.3.1) it is clear that 1

Σn

is divergent.

(ii) Consider the series, 2 2 2 2

1 1 1 1..... ......1 2 3

+ + + + +n

2

1 , 0,→∞

= =n nnu Lt u

n by p series test, clearly 2

1Σ

n converges,

Note : If 0→∞

≠nnLt u the series is divergent;

Ex. 2 12

−=

n

n nu , here 1→∞

=nnLt u ∴ Σ nu is divergent.

1.3 Tests for the Convergence of an Infinite Series In order to study the nature of any given infinite series of +ve terms regarding convergence or otherwise, a few tests are given below.

1.3.1 P-Series Test

The infinite series, 1

1 1 1 1 ......,1 2 3

∞

=Σ = + + +p p p pn n

is

(i) Convergent when p > 1, and (ii) Divergent when 1p ≤ . (JNTU 2002, 2003)

Proof :

Case (i) Let p > 1; 1,3 2> >p pp ; ⇒ 1 13 2

<p p

∴ 1 1 1 1 22 3 2 2 2

+ < + =p p p p p

Similarly, 1 1 1 1 1 1 1 1 44 5 6 7 4 4 4 4 4

+ + + < + + + =p p p p p p p p p

1 1 1 8.... ,8 9 16 8

+ + + <p p p p and so on.

Adding we get


8

1 2 4 81 ....

2 4 8Σ < + + + +p p p pn

i.e., ( ) ( ) ( )1 2 1 3 1

1 1 1 11 ......2 2 2− − −

Σ < + + + +p p p pn

The RHS of the above inequality is an infinite geometric series with common

ratio ( )1

1 1 since 12 − < >p p The sum of this geometric series is finite.

Hence 1

1∞

=Σ pn n

is also finite.

∴ The given series is convergent.

Case (ii) Let p =1; 1 1 1 11 ......2 3 4

Σ = + + + +pn

We have, 1 1 1 1 13 4 4 4 2

+ > + =

1 1 1 1 1 1 1 1 15 6 7 8 8 8 8 8 2

+ + + > + + + =

1 1 1 1 1 1 1....... .....9 10 16 16 16 16 2

+ + > + + = and so on

∴ 1 1 1 1 1 1 11 .....2 3 4 5 6 7

⎛ ⎞ ⎛ ⎞Σ = + + + + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠pn

1 1 11 .....2 2 2

≥ + + + +

The sum of RHS series is ∞

1 1since 12 2 →∞

− +⎛ ⎞= + = = ∞⎜ ⎟⎝ ⎠

n nn

n ns and Lt s

∴ The sum of the given series is also ∞ ; ∴ 1

1∞

=Σ pn n

( p = 1 ) diverges.

Case (iii) Let p<1, 1 1 11 .....2 3

Σ = + + +p p pn

Since , ,

1 1 1 11, ......2 2 3 3

< > >p pp and so on

∴ 1 1 1 11 .......2 3 4

Σ > + + + +pn

From the Case (ii), it follows that the series on the RHS of above inequality is divergent.


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∴ 1

Σ pn is divergent , when P < 1

Note: This theorem is often helpful in discussing the nature of a given infinite series.

1.3.2 Comparison Tests 1. Let Σ nu and Σ nv be two series of +ve terms and let Σ nv be convergent. Then Σ nu converges, (a) If ,≤ ∀n nu v n N∈

(b) or ≤ ∀ ∈n

n

u k n Nv

where k is > 0 and finite.

(c) or →n

n

uv

a finite limit > 0

Proof : (a) Let Σ =nv l (finite) Then, 1 2 1 2..... ...... ..... ..... 0+ + + + ≤ + + + ≤ >n nu u u v v v l Since l is finite it follows that Σ nu is convergent

(c) ,≤ ⇒ ≤ ∀ ∈nn n

n

u k u kv n Nv

, since Σ nv is convergent and k (>0) is finite,

Σ nkv is convergent ∴ Σ nu is convergent.

(d) Since →∞

n

nn

uLtv

is finite, we can find a +ve constant ,∋ < ∀ ∈n

n

uk k n Nv

∴ from (2) , it follows that Σ nu is convergent 2. Let Σ nu and Σ nv be two series of +ve terms and let Σ nv be divergent. Then

Σ nu diverges, * 1. If ,≥ ∀ ∈n nu v n N

or * 2. If ,≥ ∀ ∈n

n

u k n Nv

where k is finite and 0≠

or * 3. If →∞

n

nn

uLtv

is finite and non-zero.

Proof: 1. Let M be a +ve integer however large it may be. Science Σ nv is divergent, a

number m can be found such that 1 2 ..... ,+ + + > ∀ >nv v v M n m ∴ ( )1 2 ..... ,+ + + > ∀ > ≥n n nu u u M n m u v ∴ Σ nu is divergent


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2. 1 ≥ ∀nu kv n Σ nv is divergent ⇒ Σ nkv is divergent ∴ Σ nu is divergent

3. Since →∞

n

nn

uLtv

is finite, a + ve constant k can be found such that ,> ∀n

n

u k nv

(probably except for a finite number of terms ) ∴ From (2), it follows that Σ nu is divergent. Note :

(a) In (1) and (2), it is sufficient that the conditions with * hold ∀ > ∈n m N Alternate form of comparison tests : The above two types of comparison tests 2.8.(1) and 2.8.(2) can be clubbed together and stated as follows :

If Σ nu and Σ nv are two series of + ve terms such that n

nn

uLtv→∞

= k, where k is

non- zero and finite, then Σ nu and Σ nv both converge or both diverge.

(b) 1. The above form of comparison tests is mostly used in solving problems. 2. In order to apply the test in problems, we require a certain series Σ nv whose

nature is already known i.e., we must know whether Σ nv is convergent are divergent. For this reason, we call Σ nv as an ‘auxiliary series’.

3. In problems, the geometric series (1.2.2.) and the p-series (1.3.1) can be conveniently used as ‘auxiliary series’.

Solved Examples

EXAMPLE 1 Test the convergence of the following series:

(a) 3 4 5 6 .....1 8 27 64

+ + + + (b) 4 5 6 7 .....1 4 9 16

+ + + + (c) ( )1 44

11

∞

=

⎡ ⎤Σ + −⎢ ⎥⎣ ⎦nn n

SOLUTION

(a) Step 1: To find " "nu the thn term of the given series. The numerators 3, 4, 5, 6......of the terms, are in AP.

thn term ( )3 1 .1 2= + − = +nt n n

Denominators are 3 3 3 31 ,2 ,3 ,4 ..... thn term = 3n ; ∴ 3

2+=n

nun

Step 2: To choose the auxiliary series Σ nv . In ,nu the highest degree of n in the numerator is 1 and that of denominator is 3.


11

∴ we take, 3 1 2

1 1−= =nv

n n

Step 3: 23

2 2 21 1,→∞ →∞ →∞ →∞

+ + ⎛ ⎞= × = = + =⎜ ⎟⎝ ⎠

n

n n n nn

u n nLt Lt n Lt Ltv n n n

which is non- zero and

finite.

Step 4: Conclusion: 1→∞

=n

nn

uLtv

∴ Σ nu and Σ nv both converge or diverge (by comparison test). But 2

1Σ = Σnv

n is

convergent by p-series test (p = 2 > 1); ∴ Σ nu is convergent.

(b) 4 5 6 7 .....1 4 9 16

+ + + +

Step 1: 4 , 5, 6, 7, .....in AP , ( )4 1 1 3= + − = +nt n n ∴ 2

3+=n

nun

Step 2: Let 1Σ =nv

n be the auxiliary series

Step 3: 2

3 31 1→∞ →∞ →∞

+⎛ ⎞ ⎛ ⎞= × = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

n

n n nn

u nLt Lt n Ltv n n

, which is non-zero and finite.

Step 4: ∴ By comparison test, both Σ nu and Σ nv converge are diverge together.

But 1Σ = Σnv

n is divergent, by p-series test (p = 1); ∴ Σ nu is divergent.

(c) ( )1 44

11

∞

=

⎡ ⎤Σ + −⎢ ⎥⎣ ⎦nn n

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⎟⎠⎞

⎜⎝⎛ +=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ += 1–11–11

41

4

41

44

nnn

nn

4 8 4 8

1 1 11 1 1 34 41 . ..... 1 .....

4 2! 4 32

⎡ ⎤⎛ ⎞−⎜ ⎟⎢ ⎥ ⎡ ⎤⎝ ⎠⎢ ⎥= + + + − = − +⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦

n nn n n n

3 7 3 4

1 3 1 1 3.... .....4 32 4 32

⎡ ⎤= − + = − +⎢ ⎥⎣ ⎦n n n n

Here it will be convenient if we take 3

1=nv

n


12

4

1 1 1..... ,4 32 4→∞ →∞

⎛ ⎞= − + =⎜ ⎟⎝ ⎠

n

n nn

uLt Ltv n

which is non-zero and finite

∴ By comparison test, Σ nu and Σ nv both converge or both diverge. But by p-

series test 3

1Σ =nv

n is convergent. (p = 3 > 1); ∴ Σ nu is convergent.

EXAMPLE 2

If 3 2

34

3 12 3 5

+=

+ +n

nun n

show that Σ nu is divergent

SOLUTION

As n increases, nu approximates to

1 2 13 2 3 3 3

1 3 1 134 4 4 124

3 3 3 1.2 2 2

= × =n nn nn

∴ If we take 112

1=nv

n ,

13

14

3

2→∞=n

nn

uLtv

which is finite.

[(or) Hint: Take 1 2

1−=n l lv

n , where 1l and 2l are indices of ‘n’ of the largest terms

in denominator and nominator respectively of nu . Here 3 2 14 3 12

1 1−

= =nvnn

]

By comparison test, Σ nv and Σ nu converge or diverge together. But 112

1Σ = Σnv

n is

divergent by p – series test ( 1since 112

= <p )

∴ Σ nu is divergent.

EXAMPLE 3

Test for the convergence of the series. 1 2 3 4 ......2 3 4 5

+ + + +

SOLUTION

Here, 1

=+nnu

n; Take 1 1 0

2 2

1 1 1−

= = =nvn

n, 1 111

→∞ →∞= =

+

n

n nn

uLt Ltv

n

(finite)


13

Σ nv is divergent by p – series test. (p = 0 < 1) ∴ By comparison test, Σ nu is divergent, (Students are advised to follow the procedure given in ex. 1.2.9(a) and (b) to find “ nu ” of the given series.)

EXAMPLE 4

Show that 1 1 11 ....... .....1 2

+ + + + +n

is convergent.

SOLUTION

1=nu

n (neglecting 1st term )

= 1

1 1 11.2.3...... (2 )1.2.2.2..... 1 −< =

− nn n times

∴ 2 3

1 1 11 ......2 2 2

Σ < + + + +nu

which is an infinite geometric series with common ratio 1 12

<

∴ 1

12 −Σ n is convergent. (1.2.3(a)). Hence Σ nu is convergent.

EXAMPLE 5

Test for the convergence of the series, 1 1 1 .......1.2.3 2.3.4 3.4.5

+ + +

SOLUTION

( )( )1

1 2=

+ +nun n n

; Take 3

1=nv

n

3

31

1 21 1→∞ →∞

= =⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

n

n nn

u nLt Ltv n

n n

(finite)

∴ By comparison test, Σ nu , and Σ nv converge or diverge together. But by p-series test,

Σ nv = 3

1Σ

n is convergent ( p = 3 > 1 ); ∴ Σ nu is convergent .

EXAMPLE 6

If 4 41 1,= + − −nu n n show that Σ nu is convergent. [JNTU, 2005]

SOLUTION

1 12 22 2

4 4

1 11 1⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

nu n nn n


14

24 8 12 4 8 12

1 1 1 1 1 11 .... 1 .....2 8 16 2 8 16

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − + − − − − − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦n

n n n n n n

24 12 2 10

1 1 1 1.... 1 ....8 8

⎡ ⎤ ⎡ ⎤= + + = + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦n

n n n n

Take 2

1=nv

n , hence 1

→∞=n

nn

uLtv

∴ By comparison test, Σ nu and Σ nv converge or diverge together. But 2

1Σ =nv

n is

convergent by p –series test (p = 2 > 1) ∴ Σ nu is convergent.

EXAMPLE 7

Test the series 1 1 1 .....1 2 3

+ + ++ + +x x x

for convergence.

SOLUTION

1=

+nun x

; take 1=nv

n, then 1

1= =

+ +

n

n

u nxv n xn

1 11;1

→∞

⎛ ⎞⎜ ⎟

= Σ = Σ⎜ ⎟⎜ ⎟+⎝ ⎠

nnLt vx n

n

is divergent by p-series test (p =1 )

∴ By comparison test, Σ nu is divergent.

EXAMPLE 8

Show that 1

1sin∞

=

⎛ ⎞Σ ⎜ ⎟⎝ ⎠n n

is divergent.

SOLUTION

1sin ;⎛ ⎞= ⎜ ⎟⎝ ⎠

nun

take 1=nv

n

0

1sinsin

1→∞ →∞ →

⎛ ⎞⎜ ⎟⎝ ⎠= =

⎛ ⎞⎜ ⎟⎝ ⎠

n

n n tn

u tnLt Lt Ltv t

n

(where 1 )=t n 1=

∴ ,Σ Σn nu v both converge or diverge . But

1Σ = Σnv

n is divergent

( p -series test, 1p = ); ∴ Σ nu is divergent.


15

EXAMPLE 9

Test the series 1 1sin− ⎛ ⎞Σ ⎜ ⎟⎝ ⎠n

for convergence.

SOLUTION

1 1sin−=nun

; Take 1=nv

n

11

sin1

⎛ ⎞− ⎜ ⎟⎝ ⎠

→∞ →∞=

⎛ ⎞⎜ ⎟⎝ ⎠

nn

n nn

uLt Ltv

n

; = 1

0

11 sinsin

−

→

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Lt Takingnθ

θ θθ

But Σ nv is divergent. Hence Σ nu is divergent.

EXAMPLE 10

Show that the series 2 3

2 3 3

1 2 31 .....2 3 4

+ + + + is divergent.

SOLUTION

Neglecting the first term, the series is .....43

32

21

4

3

3

2

2 +++ . Therefore

( ) ( )( )1 1 11 1 11 . 1

+= = =+ ++ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

n n n

n n nn

n n nu nn nn n n

n n

= 11 11 1⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

n

nn n

;

Take 1=nv

n

∴ 1 1 11 1 11 1 1 .1

→∞ →∞ →∞= = =

⎛ ⎞⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

nn nn n n

n

uLt Lt Ltv e

n n n

which is finite and 1Σ = Σnv

n is divergent by p –series test ( p = 1)

∴ Σ nu is divergent.

EXAMPLE 11

Show that the series 1 3 5 .......

1.2.3 2.3.4 3.4.5+ + + ∞ is convergent. (JNTU 2000)

SOLUTION

1 3 5 .......

1.2.3 2.3.4 3.4.5+ + + ∞


16

thn term = ( )( ) 2

122 1 1 .

1 21 2 1 1n

n nun n n n

n n

⎛ ⎞−⎜ ⎟− ⎝ ⎠= =+ + ⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Take 2

1nv

n=

( )( )2 2

121 1

1 21 1n

n nn

u nLt Ltv n n

n n→∞ →∞

⎛ ⎞−⎜ ⎟ ⎛ ⎞⎝ ⎠= ÷ ⎜ ⎟⎝ ⎠+ +

( )( )

2 0 21 0 1 0

n

nn

uLtv→∞

−= =

+ + which is finite and non-zero

∴ By comparison test nu∑ and nv∑ converge or diverge together

But nv∑ = 2

1∑ n is convergent. ∴ nu∑ is also convergent.

EXAMPLE 12

Test whether the series 1

11n n n

∞

= + +∑ is convergent (JNTU 1997, 1999, 2003)

SOLUTION

The given series is 1

11n n n

∞

= + +∑

1

1nun n

=+ +

=( )( )

1 11 1

n n n nn n n n

+ −= + −

+ + + −

1

2

2

1 1 11 1 1 ..... 12 8nu n n

n n n

⎧ ⎫ ⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞= + − = + − + −⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭⎪ ⎪⎩ ⎭

⎭⎬⎫

⎩⎨⎧ +=

⎭⎬⎫

⎩⎨⎧ += ....

81–

211...

81–

21

2 nnnnnun


17

Take 1

nvn

=

1 1 2 1 1......

2 8 2n

n nn

uLt Ltv nn n→∞ →∞

⎛ ⎞⎧ ⎫= − + ÷ =⎨ ⎬ ⎜ ⎟⎩ ⎭ ⎝ ⎠

which is finite and non-zero . Using comparison test nu∑ and nv∑ converge or diverge together.

But nv∑ = 1n∑ is divergent ( )1since 2p =

∴ nu∑ is also divergent.

EXAMPLE 13

Test for convergence 3 3

1

1n

n n∞

=

⎡ ⎤+ −⎣ ⎦∑ [JNTU 1996, 2003, 2003]

thn term ( )1

3

3 3 6

1 1 11 1 13 31 1 1 . ..... 13 1.2nu n n

n n n

⎡ ⎤−⎡ ⎤⎛ ⎞ ⎢ ⎥⎢ ⎥= + − = + + + −⎜ ⎟ ⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦

= 2 5 2 3

1 1 1 1 1...... ......3 9 3 9n n n n

⎛ ⎞− + = − +⎜ ⎟⎝ ⎠

; Let 2

1nv

n=

Then ( )31 1 1.... 03 39

nn nn

uLt Ltv n→∞ →∞= − + = ≠

∴ By comparison test, nu∑ and nv∑ both converge or diverge.

But nv∑ is convergent by p -series test ( )since 2 1p = > ∴ nu∑ is convergent.

EXAMPLE 14

Show that the series, 2 3 4 .......

1 2 3p p p+ + + is convergent for p > 2 and divergent for

2p ≤

SOLUTION

thn term of the given series = ( ) ( )

1

1 11 11n p p p

nn n nun n n −

+ ++= = =

Let us take 1

1 ; 1 0;nn p n n

uv Lt vn − →∞= = ≠


18

∴ nu∑ and nv∑ both converge or diverge by comparison test.

But nv∑ = 11

pn −∑ converges when p -1>1 ; i.e., p >2 and diverges when

1 1p − ≤ i.e 2p ≤ ; Hence the result.

EXAMPLE 15

Test for convergence 1

2

1

2 33 1

n

nn

∞

=

⎛ ⎞+⎜ ⎟+⎝ ⎠

∑ (JNTU 2003)

SOLUTION

( )( )

1232 1 2

13 1 3

nn

nn

n

u⎡ ⎤+⎢ ⎥= ⎢ ⎥

+⎢ ⎥⎣ ⎦

; Take 23

n

n nv = ;

1231 2

11 3

nn

n n

uv

⎛ ⎞+⎜ ⎟=⎜ ⎟+⎝ ⎠

1 0n

nn

uLtv→∞

= ≠ ; ∴ By comparison test, nu∑ and nv∑ behave the same way.

But nv∑ =3

2 2

1

2 2 2 2 .....,3 3 3 3

n

n

∞

=

⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ which is a geometric series with

common ratio 23 (<1) ∴ nv∑ is convergent. Hence nu∑ is convergent.

EXAMPLE 16

Test for convergence of the series, 1 4 9 ......

4.7.10 7.10.13 10.13.16+ + + (JNTU 2003)

SOLUTION

4, 7, 10,..............is an A . P; ( )4 1 3 3 1nt n n= + − = +

7, 10, 13,............is an A . P; ( )7 1 3 3 4nt n n= + − = +

and 10 , 13 , 16 ,.............is an A. P; ( )10 1 3 3 7nt n n= + − = +

∴ ( )( )( ) ( ) ( ) ( )

2 2

71 43 1 3 4 3 7 3 1 .3 1 .3 13 3 3n

n nun n n n n nn n n

= =+ + + + + +

=( )( )( )

171 427 1 1 13 3 3n n n n+ + +

;


19

Taking 1

nvn

= , we get

1 027

n

nn

uLtv→∞

= ≠ ; ∴ By comparison test, both nu∑ and nv∑ behave in the same

manner. But by p –series test, nv∑ is divergent, since p = 1. ∴ nu∑ is divergent .

EXAMPLE 17

Test for convergence 2

3 2

2 5 14 7 2

n nn n

− +− +∑ (JNTU 2003)

SOLUTION

thn term of the given series = 2

3 2

2 5 14 7 2n

n nun n

− +=

− +

Let 2

1nv

n=

( )

22

33

5 12.

17 24n

n nn

nu nn nLt Ltv n n n

→∞ →∞

⎡ ⎤− +⎢ ⎥= ×⎢ ⎥− +⎢ ⎥⎣ ⎦

= ( )

2

3

5 12 2 047 24n

n nLtn n

→∞

⎡ ⎤− +⎢ ⎥ = ≠⎢ ⎥− +⎢ ⎥⎣ ⎦

∴ By comparison test, nu∑ and nv∑ both converge or diverge.

But nv∑ is convergent. [p series test – p = 2 > 1] ∴ nu∑ is convergent .

EXAMPLE 18

Test the series nu∑ , whose thn term is ( )2

14n i−

SOLUTION

( )2

14nu

n i=

−; Let 2

1nv

n= ,

( )2

22

1 044

n

n nn

u nLt Ltv in n

→∞ →∞

⎡ ⎤⎢ ⎥= = ≠⎢ ⎥

−⎢ ⎥⎣ ⎦

∴ nu∑ and nv∑ both converge or diverge by comparison test. But nv∑ is

convergent by p –series test ( )2 1p = > ; ∴ nu∑ is convergent.

Note: Test the series ∑∞

=12 141

n –n


20

EXAMPLE 19

If 1 1.sinnun n

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, show that nu∑ is convergent .

SOLUTION

Let 2

1nv

n= , so that nv∑ is convergent by p –series test .

( )

( ) 0

1sin sin1

n

n n tn

u tnLt Lt Ltv t

n→∞ →∞ →

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

where t = 1/n, Thus 1 0n

nn

uLtv→∞

⎛ ⎞= ≠⎜ ⎟

⎝ ⎠

∴ By comparison test, nu∑ is convergent.

EXAMPLE 20

Test for convergence 1 1tan( )nn∑

SOLUTION

Take 32

1nv

n= ; 1 0n

n n

uLt v→∞

⎡ ⎤ = ≠⎢ ⎥⎣ ⎦ ( as in above example)

Hence by comparison test, nu∑ converges as nv∑ converges.

EXAMPLE 21

Show that 2

1

1sinn n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

∑ is convergent.

SOLUTION

Let nu 2 1sinn

⎛ ⎞= ⎜ ⎟⎝ ⎠

; Take 2

1nv

n= ,

( ) 22

0

1sin sin1

n

n n tn

u tnLt Lt Ltv tn

→∞ →∞ →

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎝ ⎠ ⎣ ⎦

where 1t n= ; 21 1 0nn n

uLt v→∞

⎛ ⎞ = = ≠⎜ ⎟⎝ ⎠

∴ By comparison test, nu∑ and nv∑ behave the same way.

But nv∑ is convergent by p- series test, since p = 2 > 1; ∴ nu∑ is convergent.


21

EXAMPLE 22

Show that ( )2

1log n

n n

∞

=∑ is divergent.

SOLUTION 1

lognu n n= ; 1 1log 2 1 2log 2 2 2log 2 2< ⇒ < ⇒ > ;

Similarly 1 1 1 1..... ,3log 3 3, log n Nn n n> > ∈

∴ 1 1logn n n∑ > ∑ ; But 1

n∑ is divergent by p-series test.

By comparison test, given series is divergent. [If nv∑ is divergent and n nu v n≥ ∀ then

nu∑ is divergent.] (Note : This problem can also be done using Cauchy’s integral Test.

EXAMPLE 23

Test the convergence of the series ( ) ( )1

r s

nc n d n

∞− −

=

+ +∑ , where c, d, r, s are all +ve.

SOLUTION

The thn term of the series ( ) ( )

1n r su

c n d n= =

+ +.

Let 1

n r svn += Then

1

1 . 1 1 1

r sn

r s r sn r s

u nv c d c dn n

n n n n

+

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 0n

nn

uLtv→∞

= ≠ , ∴ nu∑ and nv∑ both converge are diverge, by comparison test.

But by p-series test, nv∑ converges if (r + s) > 1 and diverges if (r + s) ≤ 1

∴ nu∑ converges if ( r + s ) > 1 and diverges if ( r + s ) ≤ 1.

EXAMPLE 24

Show that ( )11

1

nn∞

− +∑ is divergent.

SOLUTION

( )11

11

.n

nn

u nn n

− += = Take

1nv

n= ; 1

1 1 0n

n n nn

uLt Ltv n→∞ →∞

= = ≠


22

For let 11

n nLt y

n→∞= say; log y =

1 .lognLt n

n→∞− =

10

1n

nLt→∞

− =

∴ 0 1y e= = (∞⎛ ⎞

⎜ ⎟∞⎝ ⎠ using L Hospitals rule)

By comparison test both nu∑ and nv∑ converge or diverge. But p-series test,

nv∑ diverges (since p =1); Hence nu∑ diverges.

EXAMPLE 25

Test for convergence the series ( )

( ) ( )1

r

p qn

n an b n c

∞

=

+

+ +∑ , a, b, c , p, q, r, being +ve.

SOLUTION

( )( ) ( )

( )( ) ( )qqpp

rr

qp

r

n

ncnn

bnn

an

cnbnanu

++

+=

+++

=11

1 =

( )( ) ( )

11 .1 1

r

p qp q r

an

n b cn n

+ −

+

+ +;

Take 1

n p q rvn + −= ; 1 0n

nn

uLtv→∞

= ≠ ;

Applying comparison tests both nu∑ and nv∑ converge or diverge.

But by p-series test, nv∑ converges if ( p+ q –r ) > 1and diverges if ( p + q –r) ≤ 1.

Hence nu∑ converges if ( p + q – r ) > 1 and diverges if ( p + q – r) ≤ 1.

EXAMPLE 26 Test the convergence of the following series whose thn terms are:

(a) ( )

( )( )( )3 4

2 1 2 3 2 5n

n n n+

+ + + ; (b)

1tann

; (c) 2

1 13

nnn n

+⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

(d) ( )

13 5n n+

; (e) 1.3nn

SOLUTION

(a) Hint : Take 2

1 ;n nv vn

= ∑ is convergent; 3 08

n

nn

uLtv→∞

⎛ ⎞= ≠⎜ ⎟

⎝ ⎠ (Verify)

Apply comparison test: nu∑ is convergent [the student is advised to work out this problem fully ]


23

(b) Proceed as in Example 8; nu∑ is convergent.

(c) Hint : Take 2

1nv

n= ;

( )

( )1

3 2

1 1 031

nnn

nn nn

u eLt Ltv e e

n→∞ →∞

+⎛ ⎞= = = ≠⎜ ⎟

⎝ ⎠ +

2

1nv

n= is convergent (work out completely for yourself )

(d) 1 1 1.

3 5 5 315

n n n n nu = =

+ ⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

; Take 15n nv = ; 1 0n

nn

uLtv→∞

⎛ ⎞= ≠⎜ ⎟

⎝ ⎠

nu∑ and nv∑ behave the same way. But nv∑ is convergent since it is a

geometric series with common ratio 1 15

<

∴ nu∑ is convergent by comparison test .

(e) 1 1 ,.3 3n n n N

n≤ ∀ ∈ , since .3 3n nn ≥ ;

∴ 1 1.3 3n nn

≤∑ ∑ …..(1)

The series on the R.H .S of (1) is convergent since it is geometric series with 1 13

r = < .

∴ By comparison test 1.3nn∑ is convergent.

EXAMPLE 27

Test the convergence of the following series.

(a) 2 2 2 2 2 2 2 2 2

1 2 1 2 3 1 2 3 411 2 1 2 3 1 2 3 4

+ + + + + ++ + +

+ + + + + + + ...............

(b) 2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3

1 2 1 2 3 1 2 3 411 2 1 2 3 1 2 3 4

+ + + + + ++ + +

+ + + + + + + ..............


24

SOLUTION

(a)

( )

( ) ( ) ( )2 2 2 2

11 2 3 .... 32

2 11 2 3 ..... 2 116

n

nnnu

nn nn n

++ + + +

= = =++ + + +

+

Take 1

nvn

= ; 3 3 0

2 1 2n

n nn

u nLt Ltv n→∞ →∞

⎛ ⎞= = ≠⎜ ⎟+⎝ ⎠

nu∑ and nv∑ behave alike by comparison test.

But nv∑ is diverges by p-series test. Hence nu∑ is divergent.

(b) ( ) ( )

( )( )

( )2 2 2

23 3 32

2 11 2 2 11 2 .... 6

1 2 ..... 3 114

n

nn n nnu

n n nnn

++ ++ + +

= = =+ + ++

Hint : Take 1

nvn

= and proceed as in (a) and show that nu∑ is divergent.

Exercise 1.1 1. Test for convergence the infinite series whose thn term is:

(a) 1

n n− [Ans : divergent]

(b) 1n nn

+ − [Ans : convergent]

(c) 2 1n n+ − [Ans : divergent]

(d) 2 1n

n − [Ans : convergent]

(e) 3 31n n+ − [Ans : divergent]

(f) ( )1

1n n + [Ans : divergent]

(g) 2 1n

n + [Ans : convergent]

(h) 3

5

2 54 1nn

++

[Ans : convergent]


25

2. Determine whether the following series are convergent or divergent.

(a) 1 2 3

1 2 31 3 1 3 1 3− − −+ + +

+ + + ............ [Ans : divergent]

(b) 3 3 3 3

12 22 32 2 10........ .....1 2 3

nn

++ + + + + ....... [Ans : convergent]

(c) 1 1 1 ......

1 2 2 3 3 4+ + +

+ + + ........ [Ans : divergent]

(d) 2 2 2

2 3 4 ......3 4 5

+ + + ......... [Ans : divergent]

(e) 2 3 4

1 1 11 2 3

+ + + .......... [Ans : convergent]

(f) 3 2

241

14 2 3n

nn n

∞

=

+

+ +∑ ............... [Ans : divergent]

(g) ( )1

1

8 1n∞

−∑ .................... [Ans : divergent]

(h) 3

51

3 85 9

nn

∞ ++∑ ........................ [Ans : convergent]

(i) 1 2 3

1.3 3.5 5.7+ + + ........... [Ans : divergent]

1.3.3 D’ Alembert’s Ratio Test

Let (i) nu∑ be a series of +ve terms and (ii) ( )1 0n

nn

uLt ku

+

→∞= ≥

Then the series nu∑ is (i) convergent if k < 1 and (ii) divergent if k > 1. Proof :

Case (i) ( )1 1n

nn

uLt ku

+

→∞= <

From the definition of a limit, it follows that

0m∃ > and ( ) 10 1 n

n

ul l l n mu

+< < ∋ < ∀ ≥


26

i.e., 1m

m

u lu

+ < , 2

1

m

m

u lu

+

+

< ,..........

∴ 1 2 ......m m mu u u+ ++ + + ........ = 1 21 .....m mm

m m

u uuu u

+ +⎡ ⎤+ + +⎢ ⎥

⎣ ⎦

1 2 1

11

m m m

mm m m

u u uu . .....

u u u

( ) ( )2 11 ... . 11m mu l l u l

l< + + + = <

−

But 1.

1mul−

is a finite quantity ∴ nn m

u∞

=∑ is convergent

By adding a finite number of terms 1 2 1...... mu u u −+ + + , the convergence of the

series is unaltered. nn m

u∞

=∑ is convergent.

Case (ii) 1 1n

nn

uLt ku

+

→∞= >

There may be some finite number of terms in the beginning which do not satisfy

the condition 1 1n

n

uu

+ ≥ . In such a case we can find a number ‘m’

1 1,n

n

u n mu

+∋ ≥ ∀ ≥

Omitting the first ‘m’ terms, if we write the series as 1 2 3u u u+ + + ........., we

have

32 4

1 2 3

1, 1, 1uu uu u u

≥ ≥ ≥ .......... and so on

∴ 32 21 2 1

1 2 1

...... 1 . .....nuu uu u u u

u u u⎛ ⎞

+ + + = + + +⎜ ⎟⎝ ⎠

(to n terms)

≥ u1 (1 + 1 + 1.1 +.......to n terms) = 1nu

11

.n

nn nn

Lt u Lt n u→∞ →∞

=

≥∑ which → ∞ ; ∴ nu∑ is divergent .


27

Note: 1 The ratio test fails when k = 1. As an example, consider the series, 1

1p

n n

∞

=∑

Here 1 1 111 1

pp

n

n n nn

u nLt Lt Ltu n n

+

→∞ →∞ →∞

⎛ ⎞⎛ ⎞ ⎜ ⎟= = =⎜ ⎟+ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

i.e., k = 1 for all values of p, But the series is convergent if p > 1 and divergent if 1p ≤ , which shows that

when k = 1, the series may converge or diverge and hence the test fails .

Note: 2 Ratio test can also be stated as follows:

If nu∑ is series of +ve terms and if 1

n

nn

uLt ku→∞

+

= , then nu∑ is convergent

If k > 1 and divergent if k < 1 (the test fails when k = 1).

Solved Examples

Test for convergence of Series

EXAMPLE 28

(a) 2 3

1.2 2.3 3.4x x x

+ + + ................. (JNTU 2003)

SOLUTION

( );

1=

+

n

nxu

n n

( )( )1

1 1 2

+

+ =+ +

n

nxu

n n;

( )( )( )1

1 1 1.21 2 1

nn

nn

n nu x xu n n x

n

++ +

= =+ + ⎛ ⎞+⎜ ⎟

⎝ ⎠

.

Therefore 1n

nn

uLt xu

+

→∞=

∴ By ratio test nu∑ is convergent When |x| < 1 and divergent when | x | > 1;

When x = 1, ( )2

11 1nu

n n=

+; Take 2

1nv

n= ; 1n

nn

uLtv→∞

=

∴ By comparison test nu∑ is convergent.

Hence nu∑ is convergent when 1x ≤ and divergent when 1x > .


28

(b) 2 31 3 5 7 .......x x x+ + + +

SOLUTION

( ) 12 1 ;−= − nnu n x ( )1 2 1+ = + n

nu n x ; 1 2 12 1

n

n nn

u nLt Lt x xu n

+

→∞ →∞

+⎛ ⎞= =⎜ ⎟−⎝ ⎠

∴ By ratio test nu∑ is convergent when 1x < and divergent when 1x >

When 1: 2 1;nx u n= = − nnLt u→∞

= ∞ ; ∴ nu∑ is divergent.

Hence nu∑ is convergent when 1x < and divergent when 1x ≥

(c) 21 1

n

n

xn

∞

= +∑ ...........

SOLUTION

2 1

n

nxu

n=

+ ;

( )

1

1 21 1

n

nxu

n

+

+ =+ +

.

Hence 2

12

12 2

n

n

u n xu n n

+ ⎛ ⎞+= ⎜ ⎟+ +⎝ ⎠

, ( )

( )2

21

22

11

2 21

n

n nn

nu nLt Lt x xu n

n n

+

→∞ →∞

⎡ ⎤+⎢ ⎥

⎢ ⎥= =⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∴ By ratio test, nu∑ is convergent when 1x < and divergent when 1x > When

2

11:1nx u

n= =

+ ; Take 2

1nv

n=

∴ By comparison test, nu∑ is convergent when 1x ≤ and divergent when 1x >

EXAMPLE 29

Test the series 2

2

1 , 01

n

n

n x xn

∞

→∞

⎛ ⎞−>⎜ ⎟+⎝ ⎠

∑ for convergence.

SOLUTION

( )( )

221

1 22

1 11 ;1 1 1

n nn n

nnu x u xn n

++

⎡ ⎤+ −⎛ ⎞−= = ⎢ ⎥⎜ ⎟+ + +⎢ ⎥⎝ ⎠ ⎣ ⎦


29

( )( )

221

2 2

12 .2 2 1

n

n nn

nu n nLt Lt xu n n n

+

→∞ →∞

⎡ ⎤+⎛ ⎞+⎢ ⎥= ⎜ ⎟+ + −⎢ ⎥⎝ ⎠⎣ ⎦

( )( )

( )( )4 2

4 2 2

1 2 1 1

1 2 2 1 –1n

n n nLt x

n n n n→∞

⎡ ⎤+ +⎢ ⎥= =

+ +⎢ ⎥⎣ ⎦

∴ By ratio test, nu∑ is convergent when 1x < and divergent when 1x > when x = 1,

2

2

11n

nun

−=

+ Take 0

1nv

n=

Applying p-series and comparison test, it can be seen that nu∑ is divergent when x = 1.

∴ nu∑ is convergent when x < 1 and divergent 1x ≥

EXAMPLE 30

Show that the series2 3 41 .....2 3 4

p p p

+ + + + , is convergent for all values of p.

SOLUTION

p

nnun

= ; ( )

1

11

p

n

nu

n+

+=

+

( )1 1

1

pn

pn nn

nu nLt Ltu n n

+

→∞ →∞

⎡ ⎤+= ×⎢ ⎥

+⎢ ⎥⎣ ⎦=

( )1 1

1

p

n

nLtn n→∞

⎧ ⎫+⎪ ⎪⎛ ⎞⎨ ⎬⎜ ⎟+ ⎝ ⎠⎪ ⎪⎩ ⎭

= ( )

1 11 0 11

p

n nLt Lt

n n→∞ →∞

⎛ ⎞× + = <⎜ ⎟+ ⎝ ⎠;

nu∑ is convergent for all ‘ p ‘ .

EXAMPLE 31

Test the convergence of the following series

1 1 1 1

1 3 5 7p p p p+ + + + ............

SOLUTION

( )

1 ;2 1

=−

n pun

( )1

12 1

+ =+

n pun


30

( )( )

( )( )

1 2 1 2 . 1 1 22 1 2 1 1 2

p pp pn

p pp pn

n n nuu n n n

+ − −= =

+ +; 1 1n

nn

uLtu

+

→∞=

∴ Ratio test fails.

Take 1

n pvn

= ; ( )

12 1 12 1

2

pn

p pn p

u nv n

n

= =− ⎛ ⎞−⎜ ⎟

⎝ ⎠

; 1 ,2

npn

n

uLtv→∞

=

which is non – zero and finite ∴ By comparison test, nu∑ and nv∑ both converge or both diverge.

But by p – series test, nv∑ = 1

pn∑ converges when p > 1 and diverges

when 1p ≤ ∴ nu∑ is convergent if p > 1 and divergent if 1p ≤ .

EXAMPLE 32

Test the convergence of the series ( ) 0;1

13 >

+∑∞

=

xn

xn

n

n

SOLUTION

( ) ( )

( )

1

1 33

1 2;

1

n n

n n

n x n xu u

n n

+

+

+ +=

+

( ) ( )

3311

32 2. . .

1 1 11nn

nn

u n n n nx xu n x n nn

++ + +⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠+

13

21 1 .1 11 1

n

n nn

u nLt Lt x xu

n n

+

→∞ →∞

⎛ ⎞+⎜ ⎟= =⎜ ⎟

⎛ ⎞⎜ ⎟+ +⎜ ⎟⎝ ⎠ ⎝ ⎠

∴ By ratio test, nu∑ converges when x < 1 and diverges when x > 1 .

When x = 1, 3

1n

nun+

=

Take 2

1nv

n= ; By comparison test nu∑ is convergent ( give proof )

∴ nu∑ is convergent if 1x ≤ and divergent if x > 1.


31

EXAMPLE 33 Test the convergence of the series (JNTU 2002)

(i) 2

21

12n

n

nn

∞

−

⎛ ⎞+⎜ ⎟

⎝ ⎠∑ (ii)

2.5.8 2.5.8.1111.5.9 1.5.9.13

+ + + ... (iii) 1 1.2 1.2.33 3.5 3.5.7

+ + +

SOLUTION

(i) 2

21

12n

n

nn

∞

−

⎛ ⎞+⎜ ⎟

⎝ ⎠∑ =

2

21 1

12n

n n

nn

∞ ∞

= =

+∑ ∑ Let 2

2

1;2n nn

nu vn

= =

( ) ( )2 2

11 1 1 2

1 1 2; .2 2

nn

n n nn

n nuuu n

++ + +

+ += =

21 1 1 1. 1 1

2 2n

n nn

uLt Ltu n

+

→∞ →∞

⎛ ⎞= + = <⎜ ⎟⎝ ⎠

∴ By ratio test nu∑ is convergent. By p –series test, nv∑ is convergent.

∴ The given series ( )n nu v+∑ ∑ is convergent.

(ii) Neglecting the first term, the series can be taken as, 2.5.8 2.5.8.111.5.9 1.5.9.13

+ +

Here, 1st term has 3 fractions ,2nd term has 4 fractions and so on .

∴ thn term contains ( n + 2 ) fractions 2. 5. 8.......are in A. P. ∴ ( )2 thn + term = 2 + ( n + 1 ) 3 = 3n + 5 ;

∴ 1. 5. 9,.......are in A. P. ∴ ( )2 thn + term = 1 + ( n + 1 ) 4 = 4n + 5

∴ ( )( )

2.5.8..... 3 51.5.9..... 4 5n

nu

n+

=+

( )( )( )( )1

2.5.8..... 3 5 3 81.5.9..... 4 5 4 9n

n nu

n n+

+ +=

+ +

( )( )

1 3 84 9

n

n

nuu n

+ +=

+ ; 1

833 1

9 44

n

n nn

nu nLt Ltu n

n

+

→∞ →∞

⎛ ⎞+⎜ ⎟⎝ ⎠= = <⎛ ⎞+⎜ ⎟⎝ ⎠

∴ By ratio test, nu∑ is convergent.


32

(iii) 1, 2, 3, ........ are in A. P thn term = n ; 3. 5. 7..........are in A.P. thn term = 2n + 1

∴ ( )

1.2.3.....3.5.7..... 2 1n

nun

⎡ ⎤= ⎢ ⎥+⎣ ⎦

( )

( )( )1

1.2.3..... 13.5.7..... 2 1 2 3n

n nu

n n+

⎡ ⎤+= ⎢ ⎥+ +⎣ ⎦

1 12 3

n

n

u nu n

+ +⎛ ⎞= ⎜ ⎟+⎝ ⎠

1

1. 11 1

3 22

n

n nn

nu nLt Ltu n

n

+

→∞ →∞

⎛ ⎞+⎜ ⎟⎝ ⎠= = <⎛ ⎞+⎜ ⎟⎝ ⎠


EXAMPLE 34

Test for convergence ( ) ( )1

1

1.3.5.... 2 1. 0

2.4.6....2n

n

nx x

n

∞−

=

−>∑ (JNTU 2001)

SOLUTION

The given series of +ve terms has ( ) 11.3.5.... 2 1

.2.4.6....2

nn

nu x

n−−

=

and ( )( )1

1.3.5.... 2 12.4.6.... 2 2

nn

nu x

n+

+=

+

1 2 12 2

n

n nn

u nLt Lt xu n

+

→∞ →∞

+⎛ ⎞= ⎜ ⎟+⎝ ⎠ =

( )( )

12 1 2 .22 1 2

n

n nLt x xn n

→∞

+=

+

∴ By ratio test, nu∑ is converges when x < 1 and diverges when x > 1 when x = 1, the

test fails.

Then ( )1.3.5.... 2 1

12.4.6.....2n

nu

n−

= < and 0nnLt u→∞

≠

∴ nu∑ is divergent. Hence nu∑ is convergent when x < 1, and divergent when 1x ≥


33

EXAMPLE 35

Test for the convergence of ( )2 12 6 2 21 ........ ..... 05 9 2 1

nn

nx x x x−⎛ ⎞−+ + + + + >⎜ ⎟+⎝ ⎠

(JNTU 2003) SOLUTION

Omitting 1st term, ( )12 2 , 22 1

nn

n nu x n−⎛ ⎞−= ≥⎜ ⎟+⎝ ⎠

and ' 'nu are all +ve.

( )( )

1

1 1

2 2

2 1

nn

n nu x

+

+ +

−=

+;

11

1

2 2 2 1. .2 1 2 2

n nn

n nn nn

uLt Lt xu

++

+→∞ →∞

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− += ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ −⎝ ⎠ ⎝ ⎠⎝ ⎠

= ( )

( )( )( )

1

11

1 12 1 2 12 2. .1 22 1 2 12 2

n nn n

n n nn n

Lt x x+

→∞ ++

⎡ ⎤− +⎢ ⎥ =⎢ ⎥+ −⎢ ⎥⎣ ⎦

;

Hence, by ratio test, nu∑ converges if x < 1 and diverges if x > 1.

When x = 1, the test fails. Then 2 2 ; 1 02 1

n

n nn nu Lt u

→∞

−= = ≠

+; ∴ nu∑ diverges

Hence nu∑ is convergent when x < 1 and divergent x > 1

EXAMPLE 36

Using ratio test show that the series ( )

0

3 4!

n

n

in

∞

=

−∑ converges (JNTU 2000)

SOLUTION

( )3 4!

n

niu n

−= ; ( ) 1

13 4

( 1)!

n

niu n

+

+−= + ; 1 3 4 0 1

1n

n nn

u iLt Ltu n

+

→∞ →∞

⎛ ⎞ −⎛ ⎞= = <⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠

Hence, by ratio test, nu∑ converges.

EXAMPLE 37

Discuss the nature of the series, ( )2 32 3 4 ........ 03.4 4.5 5.6

x x x x+ + + ∞ > (JNTU 2003)

SOLUTION Since x > 0 , the series is of +ve terms ;


34

( )( )( )

( )( )( )

11

1 22 3 3 4

n nn n

n nu x u x

n n n n+

+

+ += > =

+ + + +

( )( )( )

21 2 .

1 4n

nn

n xuLtu n n

+

→∞

⎡ ⎤+= ⎢ ⎥

+ +⎢ ⎥⎣ ⎦ =

( )

( )22 2

22

1 .;

5 41n

n

n xLt x

n n n→∞

⎡ ⎤+⎢ ⎥ =⎢ ⎥

+ +⎢ ⎥⎣ ⎦

Therefore by ratio test, nu∑ converges if x < 1 and diverges if x >1

When x = 1, the test fails; Then ( )

( )( )1

2 3n

nu

n n+

=+ +

;

Taking 1

nvn

= ; 1 0n

nn

uLtv→∞

= = ≠

∴ By comparison test nu∑ and nv∑ behave same way. But nv∑ is divergent by p-

series test. (p = 1); ∴ nu∑ is diverges when x = 1

∴ nu∑ is convergent when x < 1 and divergent when 1x ≥

EXAMPLE 38

Discuss the nature of the series ( )( )

3.6.9.....3 .54.7.10..... 3 1 3 2

nnn n+ +∑ (JNTU 2003)

SOLUTION

Here, ( ) ( )

3.6.9.....3 5 ;4.7.10..... 3 1 3 2

n

nnu

n n=

+ +

( )

( )( )( )

1

1

3.6.9.....3 3 3 5;

4.7.10..... 3 1 3 4 3 5

n

n

n nu

n n n

+

+

+=

+ + +

( )( )( )( )

1 3 2 3 3 .53 4 3 5

n

n nn

n nuLt Ltu n n

+

→∞ →∞

+ +=

+ +

= ( )( )

( )( )2

2

325.9 1 13 3 5 1549 1 13 3

n

n n nLtn n n

→∞

⎡ ⎤+ +⎢ ⎥ = >⎢ ⎥+ +⎣ ⎦

∴ By ratio test, nu∑ is divergent.


35

EXAMPLE 39

Test for convergence the series 1

1

n

nn

∞−

=∑

SOLUTION 1 n

nu n −= ; ( )1 1 nnu n −

+ = + ;

( ) n

n

n

n

n

n

n

nn

nnnn

nn

uu

⎟⎠⎞

⎜⎝⎛

+=

+=

+=+

11

)1(1–1

–1

1 1 1 1. 0. 0 111

n

n

n nn

uLt Ltu n en

+

→∞ →∞

⎛ ⎞⎜ ⎟= = = <⎜ ⎟+⎝ ⎠

∴ By ratio test nu∑ , is convergent

EXAMPLE 40

Test the series3

1

2n

nn

∞

=∑ , for convergence.

SOLUTION

32

nnun

= ; ( )3

1

2 11n

nu

n+

+=

+

( ) ( ) ( )2

3 21

3 3

112 1 11 2

n

n

n nu n nu n n n n

+++ +

= × = =+

;

1 0 1n

nn

uLtu

+

→∞= < ;


EXAMPLE 41

Test convergence of the series 2 !n

n

nn∑

SOLUTION

2 !n

n n

nun

= ; ( )

( )

1

1 1

2 1 !1

n

n n

nu

n

+

+ +

+=

+;


36

( )

( )

11

1

2 1 !. 22 ! 11

nn nn

n nn

nu n nu n nn

++

+

+ ⎛ ⎞= = ⎜ ⎟+⎝ ⎠+

( )

1 1 22 111

nnn n

n

uLt Ltu e

n

+

→∞ →∞= = <

+ (since 2 < e < 3)


EXAMPLE 42 Test the convergence of the series nu∑ where nu is

(a) 2 1

3 1n

n ++

(b) ( )

( )1

, 02 1

n

ax an

−

>+

(c) 21.2.3....

4.7.10.....3 3nn

⎛ ⎞⎜ ⎟+⎝ ⎠

(d) 1 21 3

n

n

+

+ (e)

3 2

9

3 75 11

nn n xn

⎛ ⎞+⎜ ⎟+⎝ ⎠

SOLUTION

(a) ( )2

11 2

1 1 3 13 1 1

nn

nn nn

nuLt Ltu n

++→∞ →∞

⎡ ⎤+ +⎛ ⎞ += ×⎢ ⎥⎜ ⎟ + +⎢ ⎥⎝ ⎠ ⎣ ⎦

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎠⎞⎜

⎝⎛ +

⎟⎠⎞⎜

⎝⎛ +

⎟⎠⎞⎜

⎝⎛ +

⎟⎠⎞⎜

⎝⎛ ++

∞→=

++

11

22

22

3113

3113

.11

221

nn

nn

nn

nnn

nLt

= 1 13

<


(b) ( )

( )11

2 12 3

ann

a nn nn

nu xLt Ltu xn

+−→∞ →∞

⎡ ⎤+⎛ ⎞= ×⎢ ⎥⎜ ⎟

+⎢ ⎥⎝ ⎠ ⎣ ⎦

= ( )( )

12 1 2 .32 1 2

aa a

an a a

n nLt x xn n

→∞

⎡ ⎤+⎢ ⎥ =⎢ ⎥+⎢ ⎥⎣ ⎦

By ratio test, nu∑ convergence if x < 1 and diverges if x > 1.


37

When x = 1, the test fails; Then, ( )

12 1

n aun

=+

; Taking 1

n avn

= we have,

2 1

an

n nn

u nLt Ltv n→∞ →∞

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠=

( )1 1 0

212a an

Lt

n→∞

= ≠+

and finite ( since 0a > ).

∴ By comparison test, nu∑ and nv∑ have same property

But p –series test, we have

(i) nv∑ convergent when a > 1

and (ii) divergent when 1a ≤

∴ To sum up, (i) x < 1, nu∑ is convergent a∀ .

(ii) x > 1, nu∑ is divergent a∀ .

(iii) x = 1, a > 1, nu∑ is convergent, and

(iv) x = 1, 1a ≤ , nu∑ is divergent.

(c) ( )

( )( )( )

2

1 1.2.3.... 1 4.7.10.... 3 34.7.10.... 3 3 3 6 1.2.3....

n

n nn

n n nuLt Ltu n n n

+

→∞ →∞

⎡ ⎤+ += ×⎢ ⎥+ +⎣ ⎦

( )( )

21 1 1

3 2 9n

nLt

n→∞

⎡ ⎤+= = <⎢ ⎥

+⎢ ⎥⎣ ⎦ ;

∴ By ratio test, nu∑ is convergent

(d) ( )( )

( )( )

121

11

1 2 1 3

1 3 1 2

n nn

n nn nn

uLt Ltu

+

++→∞ →∞

⎡ ⎤+ +⎢ ⎥= ×

+ +⎢ ⎥⎣ ⎦

=

12

111 2

11

1 12 1 3 122 3 1

1 1 33 1 2 13 2

n nn n

n n nn n

Lt

++

→∞ ++

⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎛ ⎞⎝ ⎠ ⎝ ⎠⎢ ⎥× = <⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎝ ⎠⎢ ⎥+ +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦



38

(e) ( ) ( )

( )

3 2 91

9 3

3 1 7 1 5 113 75 1 11

n

n nn

n nu nLt Lt xu nn

+

→∞ →∞

⎡ ⎤+ + + += × ×⎢ ⎥

++ +⎢ ⎥⎣ ⎦

= ( ) ( )

( )( )( )

3 2 93 29

9 393

111 1 5 13 1 7 1 571 3 15 1 11 3

n

nn nn n nLt xnn n n

→∞

⎡ ⎤++ + +⎢ ⎥× ×⎢ ⎥++ +⎢ ⎥⎣ ⎦

= ( ) ( )

( )( )( )

3 23 9

9

9 3939

71 1 113 1 1 5 13 511 71 3 15 1 35

n

n nn nn nLt x xnn nn n

→∞

⎡ ⎤⎧ ⎫+ + +⎨ ⎬ +⎢ ⎥⎩ ⎭⎢ ⎥× × =⎧ ⎫⎢ ⎥++ +⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦

∴ By ratio test, nu∑ converges when x < 1 and diverges when x > 1.

When x = 1, the test fails,

Then ( )

( )( )

( )3

699 9

7 73 1 133 3511 115 1 15 5

n

n n nunn n n

+ += =

+ +

Taking 6

1nv

n= , we observe that

3 05

n

nn

uLtv→∞

= ≠

∴ By comparison test and p series test, we conclude that nu∑ is convergent.

∴ nu∑ is convergent when 1x ≤ and divergent when x > 1.

Exercise – 1.2 1. Test the convergency or divergency of the series whose general term is :

(a) nx

n ............................... [Ans : 1 , 1x cgt x dgt< ≥ ]

(b) 1nnx − ........................... [Ans : 1 , 1x cgt x dgt< ≥ ]

(c) 12 22 1

nn

n x −⎛ ⎞−⎜ ⎟+⎝ ⎠

.............. [Ans : 1 , 1x cgt x dgt< ≥ ]

(d) 2

2

11

nn xn

⎛ ⎞+⎜ ⎟−⎝ ⎠

................... [Ans : 1 , 1x cgt x dgt< ≥ ]

(e) n

nn

........................ [Ans: cgt.]


39

(f) 4 .n

n

nn

.......................... [Ans: dgt.]

(g) ( )( )

3 1

3 1

n

n

n +

+ ....................... [Ans: cgt.]

2. Determine whether the following series are convergent or divergent :

(a) 2 3

1.2 3.4 5.6x x x

+ + + .............. [Ans : 1 , 1x cgt x dgt≤ > ]

(b) 2 3

2 2 212 3 4x x x

+ + + + .............. [Ans : 1 , 1x cgt x dgt≤ > ]

(c) 21

1.2.3 4.5.6 7.8.9x x

+ + + ...... [Ans : 1 , 1x cgt x dgt≤ > ]

(d) 2 3

21 ..... ....2 5 10 1

nx x x xn

+ + + + ++

[Ans : 1 , 1x cgt x dgt≤ > ]

(e) 2 3

1.2 2.3 3.4x x x

+ + + ............... [Ans : 1 , 1x cgt x dgt> ≤ ]

1.3.4 Raabe’s Test

Let nu∑ be series of +ve terms and let 1

1n

nn

uLt n ku→∞

+

⎧ ⎫⎛ ⎞⎪ ⎪− =⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

Then (i) If k > 1, nu∑ is convergent. (ii) If k < 1, nu∑ is divergent. (The test fails if k = 1)

Proof : Consider the series 1

n pvn

=∑ ∑

1

1 11 1 1 1p p

n

n

v nn n nv n n+

⎡ ⎤ ⎡ ⎤⎡ ⎤ +⎛ ⎞ ⎛ ⎞− = − = + −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

=( )

2

1 11 . .... 12

p ppnn n

⎡ ⎤−⎛ ⎞+ + + −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

= ( )1 1.

2p p

pn

−+ + ..........

1

1n

nn

vLt n pv→∞

+

⎧ ⎫− =⎨ ⎬

⎩ ⎭


40

Case (i) In this case, 1

1 1n

nn

uLt n ku→∞

+

⎧ ⎫− = >⎨ ⎬

⎩ ⎭

We choose a number ‘p’ 1k p∋ > > ; Comparing the series nu∑ with

nv∑ which is convergent , we get that nu∑ will converge if after some

fixed number of terms

1 1

1 pn n

n n

u v nu v n+ +

+⎛ ⎞> = ⎜ ⎟⎝ ⎠

i.e. If, ( )

1

1 11 .2

n

n

p pun pu n+

−⎛ ⎞− > + +⎜ ⎟ ∠⎝ ⎠

.........from (1)

i.e., If 1

1n

nn

uLt n pu→∞

+

⎛ ⎞− >⎜ ⎟

⎝ ⎠

i.e., If k > p, which is true . Hence nu∑ is convergent .The second case also

can be proved similarly.

Solved Examples EXAMPLE 43 Test for convergence the series

3 5 71 1.3 1.3.5. . . .....

2 3 2.4 5 2.4.6 7x x xx + + + + (JNTU 2006, 2008)

SOLUTION Neglecting the first tem ,the series can be taken as ,

3 5 71 1.3 1.3.5. . . .....

2 3 2.4 5 2.4.6 7x x x

+ + +

1.3.5....are in A.P. thn term = ( )1 1 2 2 1n n+ − = −

2.4.6...are in A.p. thn term = ( )2 1 2 2n n+ − =

3.5.7.....are in A.P thn term = ( )3 1 2 2 1n n+ − = +

∴ nu ( thn term of the series) ( )

( )2 11.3.5.... 2 1

.2.4.6.... 2 2 1

nn xn n

+−=

+


41

( )( )

( )( )2 3

1

1.3.5.... 2 1 2 1.

2.4.6.... 2 2 2 2 3

n

n

n n xun n n

+

+

− +=

+ +

( )( ) ( ) ( )

( )2 31

2 1

1.3.5.... 2 1 2 12.4.6....2. . .2.4.6.... 2 2 2 3 1.3.5.... 2 1

nn

nn

n nu x nu n n n x

++

+

+ +=

+ + −

= ( )

( )( )

2 22 12 2 2 3

n xn n

++ +

∴

22

2 21

2

14 12

2 34 1 12 2

n

n nn

nu nLt Lt x xu n

n n

+

→∞ →∞

⎛ ⎞+⎜ ⎟⎝ ⎠= =

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∴ By ratio test, nu∑ converges if 1x < and diverges if 1x >

If 1x = the test fails.

Then 2 1x = and ( )( )

( )21

2 2 2 32 1

n

n

n nuu n+

+ +=

+

( )( )

( ) ( )2 21

2 2 2 3 6 51 12 1 2 1

n

n

n nu nu n n+

+ + +− = − =

+ +

2

21

6 514 4 1

n

n nn

u n nLt n Ltu n n→∞ →∞

+

⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− =⎨ ⎬⎜ ⎟ ⎜ ⎟+ +⎪ ⎪ ⎝ ⎠⎝ ⎠⎩ ⎭

2

22

563 1

4 1 24n

nnLt

nn n

→∞

⎛ ⎞+⎜ ⎟⎝ ⎠= = >

⎛ ⎞+ +⎜ ⎟⎝ ⎠

By Raabe’s test, nu∑ converges. Hence the given series is convergent when 1x ≤ an

divergent when 1x > .

EXAMPLE 44 Test for the convergence of the series (JNTU 2007)

2 33 3.6 3.6.91 .....; 07 7.10 7.10.13

x x x x+ + + + >


42

SOLUTION

Neglecting the first term,

3.6.9....3 .

7.10.13....3 4n

nnu xn

=+

( )

( )( )1

1

3.6.9....3 3 3.

7.10.13.... 3 4 3 7n

n

n nu x

n n+

+

+=

+ +

1 3 3 .3 7

n

n

u n xu n

+ +=

+ ; 1n

nn

uLt xu

+

→∞=

∴ By ratio test, nu∑ is convergent when x < 1 and divergent when x > 1.

When x = 1 The ratio test fails. Then

1 1

3 7 4; 13 3 3 3

n n

n n

u unu n u n+ +

+= − =

+ +

1

4 41 13 3 3

n

n nn

u nLt n Ltu n→∞ →∞

+

⎧ ⎫⎛ ⎞⎪ ⎪ ⎛ ⎞− = = >⎨ ⎬⎜ ⎟ ⎜ ⎟+⎝ ⎠⎪ ⎪⎝ ⎠⎩ ⎭

∴ By Raabe’s test, nu∑ is convergent .Hence the given series converges if 1x ≤ and

diverges if x > 1.

EXAMPLE 45

Examine the convergence of the series ( )

( )

22 2 2

22 2 21

1 .5 .9 .... 4 34 .8 .12 .... 4n

nn

∞

=

−∑

SOLUTION

( )

( )

22 2 2

22 2 2

1 .5 .9 .... 4 34 .8 .12 .... 4n

nu

n−

= ; ( ) ( )

( ) ( )

2 22 2 2

1 2 22 2 2

1 .5 .9 .... 4 3 4 14 .8 .12 .... 4 4 4n

n nu

n n+

− +=

+

( )( )

21

2

4 11

4 4n

n nn

nuLt Ltu n

+

→∞ →∞

+= =

+ (verify)

∴ The ratio test fails. Hence by Raabe’s test, nu∑ is convergent. (give proof)


43

EXAMPLE 46

Find the nature of the series ( ) ( )

2

, 02

nnx x

n>∑ (JNTU 2003)

SOLUTION

( )2

.2

nn

nu x

n= ;

( )21

1

1.

2 2n

n

nu x

n+

+

+=

+

( )

( )( )

21 1

2 1 2 2n

n

nu xu n n

+ +=

+ +;

( )

( )( )

22

12

11.

1 2 44 1 12 2

n

n nn

nu xnLt Lt xu n n n

+

→∞ →∞

+= =

+ +

∴ By ratio test, nu∑ converges when 14

<x , i. e ; x < 4; and diverges when x >4;

When x = 4, the test fails.

( )( )

( )21

2 1 2 24 1

n

n

n nuu n+

+ +=

+

( ) ( )2

1

2 2 112 14 1

n

n

u nu nn+

− − −− = =

++;

1

11 12

n

nn

uLt nu→∞

+

⎡ ⎤⎛ ⎞ −− = <⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

∴ By ratio test, nu∑ is divergent

Hence nu∑ is convergent when x < 4 and divergent when x > 4

EXAMPLE 47

Test for convergence of the series ( )4.7.... 3 1

1.2.3....nn

xn+

∑ (JNTU 1996)

SOLUTION

( )4.7.... 3 1

1.2.3....n

n

nu x

n+

= ; ( )( )

( )1

1

4.7.... 3 1 3 41.2.3.... 1 .

nn

n nu x

n n+

+

+ +=

+

( )( )

1 3 4. 3

1n

n nn

nuLt Lt x xu n

+

→∞ →∞

⎡ ⎤+= =⎢ ⎥+⎣ ⎦


44

∴ By ratio test nu∑ converges if 3 1x < i.e., 13

x < and diverges if 13

x > ;

If 13

x = , the test fails

When 13

x = , ( )⎥⎦⎤

⎢⎣⎡

++

=⎥⎦

⎤⎢⎣

⎡

+

1–43311–

1 nnn

uu

nn

n = 1 1

43 4 3n

nn

−⎡ ⎤ = −⎢ ⎥+ ⎛ ⎞⎣ ⎦ +⎜ ⎟⎝ ⎠

1

11 13

n

nn

uLt nu→∞

+

⎡ ⎤− = − <⎢ ⎥

⎣ ⎦

∴ By Raabe’s test, nu∑ is divergent.

∴ nu∑ is convergent when 13

x < and divergent when 13

x ≥

EXAMPLE 48

Test for convergence ( )2 33 4 52 ........... 0

2 3 4x x x x+ + + + > (JNTU 2003)

SOLUTION

The thn term ( ) 11 n

n

nu x

n−+

= ; ( )( )1

21

nn

nu x

n+

+=

+;

( )( )

12

2.

1n

n

n nu xu n

+ +=

+

( )( )

2

12

2

21.

11n

n nn

nu nLt Lt x xu n n

+

→∞ →∞

+= =

+

∴ By ratio test, nu∑ is convergent if x < 1 and divergent if x > 1

If x = 1, the test fails.

Then ( )( )

2

1

11 1

2n

n nn

nuLt n Lt nu n n→∞ →∞

+

⎡ ⎤+⎡ ⎤− = −⎢ ⎥⎢ ⎥ +⎢ ⎥⎣ ⎦ ⎣ ⎦

= ( )

1 0 12n

Lt nn n→∞

⎡ ⎤= <⎢ ⎥+⎣ ⎦

∴ By Raabe’s test nu∑ is divergent

∴ nu∑ is convergent when x < 1 and divergent when 1x ≥

EXAMPLE 49

Find the nature of the series 3 3.6 3.6.9 ......4 4.7 4.7.10

+ + + ∞ (JNTU 2003)


45

SOLUTION

( )

3.6.9.....34.7.10..... 3 1n

nun

=+

; ( )

( )( )1

3.6.9.....3 3 34.7.10..... 3 1 3 4n

n nu

n n+

+=

+ +

1 3 33 4

n

n

u nu n

+ +=

+;

( )( )

133 1 3 143 1 3

n

n nn

nu nLt Ltu n n

+

→∞ →∞

+= =

+

Ratio test fails.

∴ 1

3 41 13 3

n

n nn

u nLt n Lt nu n→∞ →∞

+

⎡ ⎤⎧ ⎫ ⎡ + ⎤⎛ ⎞− = −⎨ ⎬⎢ ⎥ ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦⎩ ⎭⎣ ⎦

( ) ( )

1 113 1 33 1n n

n nLt Ltn n n

→∞ →∞= = = <

+ +

∴ By Raabe’s test nu∑ is divergent.

EXAMPLE 50 If p, q > 0 and the series

( )( )

( )( )( )( )

1.3. 1 1 21 1.3.51 ....2 2.4. 1 2.4.6 1 2

p p p p ppq q q q q q

+ + ++ + + +

+ + +

is convergent , find the relation to be satisfied by p and q.

SOLUTION

( ) ( ) ( )

( ) ( )1.3.5..... 2 1 1 ..... 1

2.4.6.....2 1 ..... 1n

n p p p nu

n q q q n− + + −

=+ + −

[neglecting 1st term]

( )( )

( )( ) ( )( )( ) ( )( )1

1.3.5..... 2 1 2 1 1 ..... 12.4.6.....2 2 2 1 ..... 1n

n n p p p n p nu

n n q q q n q n+

− + + + − +=

+ + + − +

( )( )

( )( )

1 2 12 2

n

n

n p nuu n q n

+ + +=

+ +;

( )( )

( )( )

11 12 1 2 . 112 1 12

n

n nn

pnn nu nLt Ltqu n nn n

+

→∞ →∞

⎡ ⎤++⎢ ⎥= =⎢ ⎥

+ +⎢ ⎥⎣ ⎦

∴ ratio test fails. Let us apply Raabe’s test


46

( )( )( )( )1

2 21 1

2 1n

n nn

q n nuLt n Lt nu p n n→∞ →∞

+

⎡ ⎤⎡ ⎤ ⎧ ⎫+ +⎛ ⎞ ⎪ ⎪− = −⎢ ⎥⎨ ⎬⎢ ⎥⎜ ⎟ + +⎪ ⎪⎢ ⎥⎝ ⎠ ⎩ ⎭⎣ ⎦ ⎣ ⎦

( ) ( )

( )( )2

2 1 2 111 2n

q n p n nLt n

pn n n→∞

⎡ ⎤⎧ ⎫⎢ ⎥+ − + +⎪ ⎪

⎨ ⎬⎢ ⎥⎪ ⎪⎢ + + ⎥⎩ ⎭⎣ ⎦

( ) ( )1 12 1 2 1 2 2 1

2 2n

q p q pn nLt→∞

⎡ ⎤+ − + + − +⎢ ⎥ =⎢ ⎥⎣ ⎦

Since nu∑ is convergent, by Raabe’s test, 2 2 1 1

2q p− +

>

12q p⇒ − > , is the required relation.

Exercise 1.3

1. Test whether the series 1

nu∞

∑ is convergent or divergent where

( )

( )( )

22 2 222 .4 .6 ..... 2 2

.3.4.5..... 2 1 2

nn

nu x

n n−

=−

[Ans : 1 , 1x cgt x dgt≤ > ]

2. Test for the convergence the series

( )

1

4.7.10..... 3 1 nnx

n

∞ +∑ [Ans :

1 1,3 3

x cgt x dgt< ≥ ]

3. Test for the convergence the series :

(i) 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2

2 .4 2 .4 .5 .7 2 .4 .5 .7 .8 .10 ...3 .3 3 .3 .6 .6 3 .3 .6 .6 .9 .9

+ + + [Ans : divergent]

(ii) ( )2 33.4 4.5 5.6 ..... 01.2 2.3 3.4

x x x x+ + + > [Ans : cgt if x ≤ 1 dgt if , x > 1]

(iii) ( )

( ) ( )1.3.5..... 2 1. 0

2.4.6.....2 2 2

nn x xn n−

>+∑ [Ans : cgt if x ≤ 1 dgt if , x > 1]


47

(iv) ( ) ( ) ( ) ( )

2 2 22 31 2 31 ...... 0

2 4 6x x

x x+ + + + >

[Ans : cgt if x 4< and dgt if , x 4≥ ] 1.3.5 Cauchy’s Root Test

Let nu∑ be a series of +ve terms and let1

nnn

Lt u l→∞

= . Then nu∑ is convergent when

l < 1 and divergent when l > 1

Proof : (i) 1

1nnn

Lt u l a→∞

= < ⇒ ∃ +ve number ( )1

' ' 1 ,nnl u n mλ λ λ< < ∋ < ∀ >

(or) ,nnu n mλ< ∀ >

Since 1, nλ λ< ∑ is a geometric series with common ratio < 1 and

therefore convergent. Hence nu∑ is convergent.

(ii) 1

1nnn

Lt u l→∞

= >

∴ By the definition of a limit we can find a number 1

1nnr u n r∋ > ∀ >

i.e., nu n r> ∀ >

i.e., after the 1st ‘r ‘ terms , each term is > 1. nn

Lt u→∞

= ∞∑ ∴ nu∑ is divergent.

Note : When ( )11,n

nn

uLt→∞

= the root test can’t decide the nature of nu∑ . The fact of

this statement can be observed by the following two examples.

1. Consider the series 31

13 13

1 11 : 1n

nn

nn n nun n n

Lt Lt Lt→∞ →∞ →∞

⎛ ⎞⎛ ⎞− = = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

∑

2. Consider the series 1 ,n∑ in which 1

11 1n

nnn n

un

Lt Lt→∞ →∞

= =

In both the examples given above, 1

1nn

nuLt

→∞

= . But series (1) is convergent

(p-series test) And series (2) is divergent. Hence when the limit=1, the test fails.


48

Solved Examples EXAMPLE 51 Test for convergence the infinite series whose thn terms are:

(i) 2

1nn

(ii) 1

(log )nn (iii) 2

1

11n

n⎡ ⎤+⎢ ⎥⎣ ⎦

(JNTU 1996, 1998, 2001)

SOLUTION

(i) 1

2 2

1 1, nn nnu u

n n= = ;

1

2

1 0 1;nnn n

Lt u Ltn→∞ →∞

= = <

By root test nu∑ is convergent.

(ii) 11 1;

(log ) logn

n nnu un n

= = ; 1 1 0 1;

logn

nn nLt u Lt

n→∞ →∞= = <

∴ By root test , nu∑ is convergent.

(iii) 2

11 1;11 11

nn n nn

u u

nn

= =⎛ ⎞⎛ ⎞ ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

1 1 1 1;

11→∞ →∞

= = <⎛ ⎞+⎜ ⎟⎝ ⎠

nn n

n nu

en

Lt Lt

∴ By root test nu∑ is convergent.

EXAMPLE 52 Find whether the following series are convergent or divergent.

(i) 1

13 1n

n

∞

= −∑ (ii) 2 3 4

1 1 11 .....2 3 4

+ + + + (iii) ( )

11

1n

nn

n xn

∞

+=

+⎡ ⎤⎣ ⎦∑

SOLUTION

(i)

1

11 1 1

13 1 3 13

n

nn

n nn

n

u

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟= =⎜ ⎟− ⎛ ⎞⎜ ⎟⎝ ⎠ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠


49

1

1 1 1 11 33 13

n

nnn n n

n

Lt u Lt→∞ →∞

⎛ ⎞⎜ ⎟⎜ ⎟= = <

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

; By root test, nu∑ is convergent.

(ii) 1

11 1; 0 1n

nn nn nn n

u Lt u Ltn n→∞ →∞

⎛ ⎞= = = <⎜ ⎟⎝ ⎠

; By root test, nu∑ is convergent.

(iii) ( )

1

1n

n n

n xu

n +

+⎡ ⎤⎣ ⎦=

( ){ }

1

1

1

1n n

nn nn n

n xLt u Lt

n +→∞ →∞

⎡ ⎤+⎢ ⎥=⎢ ⎥⎣ ⎦

( )

1

1 1.n n

n

n xLt

n n→∞

⎡ ⎤+⎧ ⎫⎢ ⎥⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦

= 11 1.

n n

nLt xn n→∞

+⎛ ⎞⎜ ⎟⎝ ⎠

11 11 .

n nLt x

n n→∞

⎛ ⎞+⎜ ⎟⎝ ⎠

= 11.

n nLt x x

n→∞= 1

1since . 1n nLt x

n→∞

⎛ ⎞=⎜ ⎟

⎝ ⎠

∴ nu∑ is convergent if 1x < and divergent if 1x > and when 1x = the test fails.

Then ( )

1

1 n

n n

nu

n +

+= ; Take

1nv

n=

( ) ( )

1

1 1 1. 1n n n

nn n

n

n nu nv n n n+

+ + ⎛ ⎞= = = +⎜ ⎟⎝ ⎠

; 1n

nn

uLt ev→∞

= >

∴ By comparison test, nu∑ is divergent.

( nv∑ diverges by p –series test )

Hence nu∑ is convergent if 1x < and divergent 1x ≥

EXAMPLE 53

If ( )

2

2

1n

n

nnu

n=

+, show that nu∑ is convergent.


50

( )

2

2

1

1

1n

nn

nnn n

nLt u Ltn→∞ →∞

⎡ ⎤⎢ ⎥=⎢ ⎥+⎣ ⎦

; = ( ) 11

n

nn

n n

n nLt Ltnn→∞ →∞

⎛ ⎞= = ⎜ ⎟+⎝ ⎠+

= 1 1 111

n

nLt

en

→∞

⎛ ⎞⎜ ⎟

= <⎜ ⎟⎜ ⎟+⎝ ⎠

; ∴ nu∑ converges by root test .

EXAMPLE 54

Establish the convergence of the series 2 31 2 3

3 5 7⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

........

SOLUTION

2 1

n

nnu

n⎛ ⎞= ⎜ ⎟+⎝ ⎠

.........(verify); 1 1 1

2 1 2n

nn n


⎛ ⎞= = <⎜ ⎟+⎝ ⎠

By root test, nu∑ is convergent.

EXAMPLE 55

Test for the convergence of 1

.1

n

n

n xn

∞

= +∑

SOLUTION

12

1 .11

nnu x

n

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟+⎝ ⎠

;

12

1 1 .11n

nn nLt u Lt x x

n→∞ →∞

⎛ ⎞⎜ ⎟

= =⎜ ⎟⎜ ⎟+⎝ ⎠

∴ By root test, nu∑ is convergent if 1x < and divergent if 1x > .

When 1x = : 1n

nun

=+

, taking 0

1nv

n= and applying comparison test , it can be

seen that is divergent

nu∑ is convergent if 1x < and divergent if 1x ≥ .

EXAMPLE 56

Show that ( )1

11

nn

nn

∞

=

−∑ converges.


51

SOLUTION

( )11

nn

nu n= −

( ) ( )1 1 11 1 1 0 1 since 1n n n

nn n nLt u Lt n Lt n→∞ →∞ →∞

= − = − = < = ;

∴ nu∑ is convergent by root test.

EXAMPLE 57

Examine the convergence of the series whose thn term is 2 .3

nnn x

n+⎛ ⎞

⎜ ⎟+⎝ ⎠

SOLUTION

2 .3

nn

nnu xn

+⎛ ⎞= ⎜ ⎟+⎝ ⎠;

1 23

nnn n

nLt u Lt x xn→∞ →∞

+⎛ ⎞= =⎜ ⎟+⎝ ⎠

∴ By root test, nu∑ converges when 1x < and diverges when 1x > .

When 1x = : 23

n

nnun

+⎛ ⎞= ⎜ ⎟+⎝ ⎠;

21

31

n

n nn n

nLt u Lt

n

→∞ →∞

⎛ ⎞+⎜ ⎟⎝ ⎠=⎛ ⎞+⎜ ⎟⎝ ⎠

= 2

3

1 0ee e

= ≠ and the terms are all +ve .

∴ nu∑ is divergent . Hence nu∑ is convergent if 1x < and divergent if 1x ≥ .

EXAMPLE 58

Show that the series,

1 2 32 3 4

2 3 4

2 2 3 3 4 4 ......1 1 2 2 3 3

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

− + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

is convergent (JNTU 2002)

( ) 1

1

1 1nn

n n

n nun n

−+

+

⎡ ⎤+ += −⎢ ⎥

⎢ ⎥⎣ ⎦; =

1 1 1nn nn n

n n

−− ⎡ ⎤+ +⎛ ⎞ ⎛ ⎞ −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

1 11 1 1

nn n

n n

−− ⎡ ⎤⎛ ⎞ ⎛ ⎞+ + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

;

111 1 11 1 1

nn

nun n

−− ⎡ ⎤⎛ ⎞ ⎛ ⎞= + + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦


52

1 1

1 11 1 1n

n n

=⎛ ⎞ ⎧ ⎫⎪ ⎪⎛ ⎞+⎜ ⎟ + −⎨ ⎬⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

∴ 1 1 1.

1 1n

nnLt u

e→∞=

−=

1 11e

<−

∴ By root test, nu∑ is convergent.

EXAMPLE 59

Test 1

mm

u∞

=∑ for convergence when

( )2

21

m

m m

eu

m

−

−=

+

SOLUTION

( ) ( )2

1

121

mm

mm mm m

mLt u Lte→∞ →∞

⎡ ⎤+⎢ ⎥= ⎢ ⎥

⎢ ⎥⎣ ⎦

; 21 21 1

m

m

eLt ee m e→∞

⎛ ⎞+ = = >⎜ ⎟⎝ ⎠

Hence Cauchy’s root tells us that mu∑ is divergent.

EXAMPLE 60

Test the convergence of the series 2 .n

ne

∑

SOLUTION

1

10 1

nn

n nn n

nLt u Lte→∞ →∞

= = < ∴ By root test, nu∑ is convergent.

EXAMPLE 61

Test the convergence of the series, ( )2

22 3 1

1 .2 3 ... ......, 01 2

n n

n

n xx x x

n +

++ + + >

SOLUTION

( )

1

1

1

1 .n nnn

n nn n

n xLt u Lt

n +→∞ →∞

⎡ ⎤+= ⎢ ⎥

⎢ ⎥⎣ ⎦ 1

1 1. .n n

nLt xn n→∞

⎡ ⎤+⎛ ⎞= ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦


53

11 11 . .

n nLt x

n n→∞

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

1

1.1. since 1nn

x x Lt n→∞

⎡ ⎤= = =⎢ ⎥⎣ ⎦

∴ By root test, nu∑ converges if 1x < and diverges when 1x > . When 1x = , the test fails.

Then 1 11 .

n

nun n

⎛ ⎞= +⎜ ⎟⎝ ⎠

; Take 1

nvn

=

11 0

nn

n nn

uLt Lt ev n→∞ →∞

⎛ ⎞= + = ≠⎜ ⎟⎝ ⎠

∴ By comparison test and p-series test, nu∑ is divergent.

Hence nu∑ is convergent when 1x < and divergent when 1x ≥ .

Exercise 1.4 1. Test for convergence the infinite series whose thn terms are:

(a) 1

2 1n − ............................. [Ans : convergent]

(b) ( )

( )2

1 . 1log n n ≠ ...................... [Ans : convergent]

(c) 3 1 .4 3

nn xn

+⎛ ⎞⎜ ⎟+⎝ ⎠

................................. [Ans : 4 4,3 3

x cgt x dgt< ≥ ]

(d) nxn

............................................ [Ans : cgt for all 0x ≥ ]

(e) n

nn

............................................. [Ans : convergent]

(f) 3

3 .n nn∠

........................................ [Ans : convergent]

(g) ( )

( )

2

2

2 1

2

n

n

n

n

− ................................... [Ans : convergent]

(h) ( )211

nnn − ................................... [Ans : convergent]


54

(i) 2

1 nnn

−−⎛ ⎞⎜ ⎟⎝ ⎠

................................... [Ans : divergent]

(j) ( ), 01

nnx xn

⎛ ⎞ >⎜ ⎟+⎝ ⎠ ......................... [Ans : 1x < cgt , 1x ≥ dgt]

2. Examine the following series for convergence :

(a) 2 3

2 31 ....., 02 3 4x x x x+ + + + > .................. [Ans : 1 , 1x cgt x dgt≤ > ]

(b) 2 31 2 3 .....

4 7 10⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

....................... [Ans : convergent]

1.3.6 Integral Test +ve term series,

( ) ( ) ( )1 2 ..... .....nφ φ φ+ + + +

where ( )nφ decreases as n increases is convergent or divergent according as the

integral ( )1

x dxφ∞

∫ is finite or infinite.

Proof : Let ( ) ( ) ( )1 2 .....nS nφ φ φ= + + +

From the above figure, it can be seen that the area under the curve y = ( )xφ between any two ordinates lies between the set of exterior and interior rectangles formed by the ordinates at

n = 1, 2, 3, .....n, n +1 ,...... Hence the total area under the curve lies between the sum of areas of all interior rectangles and sum of the areas of all the exterior rectangles.

Hence

( ) ( ) ( ){ } ( ) ( ) ( ) ( ){ }1

11 2 ..... 2 3 ..... 1

nn x dx nφ φ φ φ φ φ φ

++ + + ≥ ≥ + + + +∫

∴ ( ) ( )111n nS x dx Sφ φ

∞

+≥ ≥ −∫


55

1 2 3 n1A 2A 3A nA A 1nn +

1B

2B

3B

1nB -

nB1nB +

O1

O2O

3On O 1n +

1P

2P

3P

1c

2c

3c

nc1nc +

OY= ( )n

As ,n → ∞ Lt nS is finite or infinite according as ( )

1x dxφ

∞


Hence the theorem.

Solved Examples EXAMPLE 62

Test for convergence the series 2

1logn n n

∞

=∑ (JNTU 2003)

SOLUTION

2 2

1 1log log

n

ndx Lt dx

x x x x∞

→∞

⎡ ⎤= ⎢ ⎥

⎣ ⎦∫ ∫ = [ ]2

log log n

nLt x→∞

= ∞

∴ By integral test, the given series is divergent.

EXAMPLE 63

Test for convergence of the series 1

1p

n n

∞

=∑ (JNTU 2003)

SOLUTION

1 1

1 1n

p pndx Lt dx

x x∞

→∞

⎡ ⎤= ⎢ ⎥⎣ ⎦∫ ∫ = 1

11

np

n

xLtp

− +

→∞

⎡ ⎤⎢ ⎥− +⎣ ⎦

;

= 11 11

p

nLt n

p−

→∞⎡ ⎤−⎣ ⎦−

Case (i) If p >1, this limit is finite; ∴ 1

pn∑ is convergent.


56

Case (ii) If p < 1, the limit is in finite; ∴ 1

pn∑ is divergent.

Case (iii) If p = 1, the limit ( )1log logn

n nLt x Lt n→∞ →∞

= = ∞ ; ∴ 1

pn∑ is divergent.

Hence 1

pn∑ is convergent if p > 1 and divergent if p ≤ 1

EXAMPLE 64

Test the series 2

1n

ne

∞

∑ for convergence.

SOLUTION

( )( )2n n

nu n saye

φ= = ;

( )nφ is +ve and decreases as n increases. So let us apply the integral test.

( ) 2

1 1

xx dx xe dxφ∞ ∞

−=∫ ∫ = { }2

1

1 , 22

te dt t x dt xdx∞

− = =∫

= 11 1 1 102 2 2

tee e

− ∞ ⎛ ⎞− = − − =⎜ ⎟⎝ ⎠

, which is finite.

By integral test, nu∑ is convergent.

EXAMPLE 65

Apply integral test to test the convergence of the series 22

1 sinn n

π∞ ⎛ ⎞⎜ ⎟⎝ ⎠

∑

SOLUTION

Let ( ) 2

1 sinnn n

πφ ⎛ ⎞= ⎜ ⎟⎝ ⎠

; ( )nφ decreases as n increases and is +ve .

( ) 22 2

1 sinx dx dxx x

πφ∞ ∞ ⎛ ⎞= ⎜ ⎟

⎝ ⎠∫ ∫ ; Let txπ =

00

22

1 1 1sin costdt t πππ π π

− = =∫ finite, 2 dx dtxπ− = ; 2

1 1dx dtx π= −

∴ By integral test, nu∑ converges 2 2x t π= ⇒ = 0x t= ∞ ⇒ =


57

EXAMPLE 66

Apply integral test and determine the convergence of the following series.

(a) 21

34 1

nn

∞

+∑ (b) 3

41

23 2

nn

∞

+∑ (c) 1

13 1

∞

+∑ n

SOLUTION

(a) ( ) 2

34 1

nnn

φ =+

is +ve and decreases as n increases

( ) 21 1

34 1

xx dx dxx

φ∞ ∞

=+∫ ∫

2 14 1 81 5,

x t xdx dt

x t x t

⎛ ⎞+ = ⇒ =⎜ ⎟⎜ ⎟= ⇒ = = ∞ ⇒ = ∞⎝ ⎠

( )1 5

3 3 log t log 58 8

t

n n

dtx dx Lt Ltt

φ∞

→∞ →∞

⎡ ⎤ ⎡ ⎤= = − = ∞⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦∫ ∫

∴ By integral test, nu∑ diverges.

(b) ( )3

4

23 2

nnn

φ =+

decreases as n increases and is +ve

( )3

41 1

23 2

xx dx dxx

φ∞ ∞

=+∫ ∫

= [ ]55

1 1 log6 6

dt tt

∞∞= = ∞∫ [where t 43 2x= + ]

By integral test, nu∑ is divergent.

(c) ( ) 13 1

nn

φ =+

is +ve, and decreases as n increases.

( ) [ ]1 1 4

1 1 13 1 log3 1 3 3 t

dtx dx dx t x tx t

φ∞ ∞ ∞

∞= = = + = = ∞+∫ ∫ ∫

∴ By integral test, nu∑ is divergent.

Alternating Series A series, ( ) 1

1 2 3 4 ..... 1 .....nnu u u u u−− + − + − + + − + , where nu are all +ve, is an

alternating series.


58

1.3.7 Leibneitz Test

If in an alternating series ( ) 1

11 .n

nn

u∞

−

=

−∑ , where nu are all +ve,

(i) 1, ,n nu u n+> ∀ and (ii) 0nnLt u→∞

= , then the series is convergent.

Proof : Let 1 2 3 4 ......u u u u− + − + be an alternating series (‘ nu ’are all +ve)

Let 1 2 3 4......u u u u> > > , Then the series may be written in each of the following two forms :

( ) ( ) ( )1 2 3 4 5 6 .....u u u u u u− + − + − + .........(A)

( ) ( )1 2 3 4 5 .....u u u u u− − − − − ..........(B) (A) Shows that the sum of any number of terms is +ve and (B) Shows that the sum of any number of terms is 1u< . Hence the sum of the series is finite. ∴ The series is convergent.

Note : If 0nnLt u→∞

≠ , then the series is oscillatory.

Solved Examples

EXAMPLE 67

Consider the series 1 1 11 .....2 3 4

− + −

In this series, each term is numerically less than its preceding term and thn term 0→ as n → ∞ .

∴ By Leibneitz’s test, the series is convergent.

(Note the sum of the above series is 2eLog )

EXAMPLE 68

Test for convergence ( ) 112 1

n

n

−−−∑ (JNTU 1997)

SOLUTION

The given series is an alternating series ( ) 11 nnu−−∑ , where

12 1nu

n=

−

We observe that (i) 0,nu n> ∀ (ii) 1,n nu u n+> ∀ (iii) 0nnLt u→∞

=

∴ By Leibneitz’s test, the given series is convergent.


59

EXAMPLE 69

Show that the series 1 1 11 ......3 9 27

S = − + − + converges. (JNTU 2000)

SOLUTION

The given series is ( ) ( )

11

11

11

3

nn

nn u−∞

−

−

−= −∑ ∑ , where 1

13n nu −= is an alternating series

in which 1. 0,nu n> ∀ 2. 1,n nu u n+> ∀ and 3. 0nnLt u→∞

= ;

Hence by Leibneitz’s test, it is convergent.

EXAMPLE 70

Test for convergence of the series, 2 3

2 3 .....,0 11 1 1

x x x xx x x

− + − + < <+ + +

(JNTU 2003) SOLUTION

The given series is of the form( ) ( )

111 .

11

n nn

nn

xu

x

−−−

= −+∑ ∑ ,

where 1

n

n n

xux

=+

Since 0 < x < 1, 0,nu n> ∀ ;

Further, 1

1 11 1

n n

n n n n

x xu ux x

+

+ +− = −+ +

( )( )

( )( )( )

1

1 1

11 1 1 1

nn n

n n n n

x xx xx x x x

+

+ +

−−= =

+ + + + .

0 < x < 1 ⇒ all terms in numerator and denominator of the above expression are +ve. ∴ 1, .n nu u n+> ∀

Again, 0nx → as nx → ∞ since 0 < x < 1; ∴ 0 0

1 0nnLt u→∞

= =+

∴ By Leibneitz’s test, the given series is convergent.

EXAMPLE 71

Test for convergence ( )

( )( )

1

2

1

1 2

n

n n n n

−∞

=

−

+ +∑ (JNTU 2004)


60

SOLUTION The given series is an alternating series ( ) 11 n

nu−−∑

where ( )( )

11 2

nun n n

=+ +

; 0,nu n> ∀ ;

Again, ( )( )( ) ( )( )1 2 3 1 2n n n n n n+ + + > + +

∴ ( )( )( ) ( )( )

1 1 ,1 2 3 1 2

nn n n n n n

< ∀+ + + + +

i.e., 1 ,n nu u n+ < ∀

Further, ( )( )

1 01 2

nn nLt u Lt

n n n→∞ →∞= =

+ +

∴ By Leibnitz’s test, ( ) 1

21 n

nu∞

−−∑ is convergent

EXAMPLE 72 Test for the convergence of the following series,

1 2 3 4 5 ....6 11 16 21 26

− + − + − + (JNTU 1998, 2004)

SOLUTION

Given series, ( ) ( )1 1

11 1

5 1n n

nn

n un

∞− −

=

− = −+∑ ∑ is an alternating series

05 1n

nu nn

= > ∀+

; ( )( ) 1

1 1 ,5 1 5 6 5 1 5 6 n n

n n u u nn n n n +

+ −− = ⇒ < ∀

+ + + +

Again, 1 0

5 1 5nn n


= = ≠+

Thus conditions (ii) or (iii) of Leibnitz’s test are not satisfied. The given series is not convergent. It is oscillatory.

EXAMPLE 73 Test the nature of the following series.

(a) ( ) 1

1

11

n

n n

−∞ −

+ +∑ (b)

( ) 1

2

11

n

n

−−+∑ (c)

( ) 1

1

11

n

n n

−∞

=

−+∑


61

SOLUTION

(a) 1 0

1nu nn n

= > ∀+ +

;

11 1

1 1 2n nu un n n n+− = −

+ + + + +

( ) ( ) ( )( )( )2 2 0

1 1 2 2 1 1 2n n

n n n n n n n n n n+ −

= = >+ + + + + + + + + + + +

∴ By Leibnitz’s test the series converges.

(b) 2

1 0,1nu n

n= > ∀

+;

( ) 122

1 1 ,1 1 1

n nu u nn n

+> ⇒ > ∀+ + +

;

0nnLt u→∞

= ∴ By Leibnitz’s test, given series converges.

(c) 1 0,

1nu nn

= > ∀+

;

11 12 1 ,

2 1 n nn n u u nn n ++ > + ⇒ < ⇒ > ∀

+ +

By Leibnitz’s test, given series converges.

EXAMPLE 74

Test the convergence of the series 1 1 1 ......

5 2 5 3 5 4− + − + (JNTU 2004)

SOLUTION

The series can be written as ( ) 1

1

15 1

n

n n

−∞

=

−

+∑ ;

15 1nu

n=

+

(i) 0nu n> ∀

(ii) 11 15 2 5 1

5 2 5 1 n nn n u u nn n ++ > + ⇒ < ⇒ > ∀

+ +

(iii) 0nnLt u→∞

= ;

By Leibnitz’s test, the given series converges.

EXAMPLE 75 Test for convergence the series, 1 1 11 .....2 4 6− + − + (JNTU 1997)


62

SOLUTION

The given series can be written as ( )1

2

n

n−

∑ (omitting 1st term)

1 02

nn

> ∀ ; 11 1 ,2 2 2 n nu u n

n n +> ⇒ > ∀+

; 1 02n

Ltn→∞

=

∴ By Leibnitz’s test, ( )1

2

n

n−

∑ is convergent.

EXAMPLE 76

Test for convergence the series, 1 1 11 ......3! 5! 7!

− + − + (JNTU 2004, 2007)

SOLUTION

General term of the series is ( )

( )

112 1 !

n

n

−−−

The series is an alternating series; ( )

1 02 1 !

nn

> ∀−

( ) ( ) 1

1 1 ,2 1 ! 2 1 ! n nu u n N

n n +> ⇒ > ∀ ∈− −

; ( )

1 02 1 !n

Ltn→∞

=−

By Leibnitz’s test, given series is convergent.

1.4 Absolute convergence A series nu∑ is said to be absolutely convergent if the series | |nu∑ is convergent Ex. Consider the series

3 3 3

1 1 11 ......2 3 4nu = − + − + − +∑

3 3 3 31

1 1 1 11 ......2 3 4nu

n

∞

= + + + + =∑ ∑

By p − series test, nu∑ is convergent ( 3 1p = > )

Hence nu∑ is absolutely convergent.

Note: 1. If nu∑ is a series of +ve terms, then nu∑ = | |nu∑ . For such a series, there is no difference between convergence and absolute convergence. Thus a series of +ve terms is convergent as well as absolutely convergent.


63

2. An absolutely convergent series is convergent. But the converse need not be true.

Consider 1

1

1 1 1 1( 1) . 1 ......2 3 4

n

n

∞−− = − + − +∑

This series is convergent (1.7.3)

But 1 1 1 11( 1) . 1 .......2 3 4

n

n−− = + + + +∑ is divergent (p-series test).

Thus nu∑ is convergent need not imply that | |nu∑ is convergent (i.e.,

nu∑ is not absolutely convergent).

1.5 Conditional Convergence If the series nu∑ is divergent and nu∑ is convergent, then nu∑ is said to be conditionally convergent. Ex. Consider the Series

1 1 11 .......2 3 4

− + − nu∑ is convergent by Leibnitz’s test. (Ex.1.7.3)

But 1 1 11 .....2 3 4nu = + + +∑ is divergent by p − series test.

∴ nu∑ is conditionally convergent.

1.6 Power Series and Interval of Convergence A series, 2

0 1 2 ..... .....nna a x a x a x+ + + + + where ‘ na ’ are all constants is a power

series in x .

It may converge for some values of x .

1 1 .n n

n nn n

u aLt Lt xu a

+ +

→∞ →∞= (1st term is omitted.) = kx (say) where 1n

nn

aLt ka

+

→∞=

Then, by ratio test, the series converges when 1kx < .

i.e., it converges 1 1, ( 0)x k

k k−⎛ ⎞∀ ∈ ≠⎜ ⎟

⎝ ⎠


64

The interval 1 1,

k k−⎛ ⎞

⎜ ⎟⎝ ⎠

is known as the interval of convergence of the given power

series.

Solved Examples EXAMPLE 77

Find the interval of convergence of the series 31

n

n

xn

∞

=∑

SOLUTION

1

13 3;( 1)

n n

n nx xu un n

+

+= =+

3

31 1. .11 1

n

n n nn

u nLt Lt x Lt x xu n

n

+

→∞ →∞ →∞

⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠ ⎜ ⎟+⎝ ⎠

By ratio test, the given series converges when 1,x < i.e., ( 1,1)x ∈ −

When 311, ,nx u n= =∑ ∑ which, is convergent by p series test.

∴ nu∑ is convergent when 1x = Hence, the interval of convergence of the given series is (–1, 1)

EXAMPLE 78 Test for the convergence of the following series. (a) 1 1 11 ..........

2 3 4− + − + (JNTU 1996)

(b) 2 2 2 2 2 2 21 1 1 1 1 1 11 .......2 3 4 5 6 7 8+ − − + + − − + (JNTU 1998)

(c) 2 4 6

1 ..........2 4 6x x x− + − + (JNTU 2004)

(d) 0

( 1) ( 1) ,n nn x∞

− +∑ with 12

x < (JNTU 2004)

SOLUTION (a) The series is of the form 1( 1)n

nu−−∑ where 1nu

n=


65

It is an alternating series where (i) 0n nu > ∀ (ii) 1n nu u n+> ∀ and (iii) 0;nn

Lt u→∞

= ∴ By Leibnitz test, the series is convergent.

Again the series 1 1 11 ......2 3 4

+ + + + is divergent, by

p − series test. Hence the given series is conditionally convergent.

(b) 21

1n

p

u n

∞

=

=∑ ∑ which is convergent by p − series test .

∴ The given series is absolutely convergent. ∴ It is convergent. (c) The given series is

2 2 2 2

1 1( 1) . . = (-1) ; (2 2)! (2 2)!

n nn n

n nx xu un n

− −− −− ∴ =

− −∑ ∑

2

1 ;2 !

n

nxun+ = 21 1 .

(2 1)(2 )n

n

u xu n n

+ =−

; 1 0 1n

nn

uLtu

+

→∞= <

By ratio test, the series nu∑ converges ;x∀ i.e., nu∑ is absolutely

convergent ;x∀ (d) Here, 1

1( 1) ;| | ( 2)n nn nu n x u n x +

+= + = + (neglect 1st term)

12(1 )( 2) 11( 1) (1 )

n

n n nn

u n nLt Lt x Lt x xu n n

+

→∞ →∞ →∞

++= = = <

+ + 1( )2x <∵

∴ nu∑ is convergent x∀ , i.e., given series is absolutely convergent and hence convergent.

EXAMPLE 79 Show that the series

2 21 .....2 3

x xx+ + + + converges absolutely x∀

SOLUTION

1 0 1n

nn

u xLt

u n+

→∞= = < when 0x ≠ [since

1

1;( 1)! !

n n

n n

x xu u

n n

−

+= =−

]

∴ By ratio test, nu∑ is convergent 0x∀ ≠ .


66

When 0x = , the series is (1 + 0 + 0 + ......) and is convergent

∴ nu∑ converges nu⇒ ∑ is absolutely convergent x∀ .

EXAMPLE 80

Show that the series, 2 4

1 1 11 ........3 3 3

− + − + is absolutely convergent.

SOLUTION

11

13nn

nu

∞

−=

=∑ ∑ , which is a geometric series with common ratio 1 13

<

∴ It is convergent. Hence given series is absolutely convergent.

EXAMPLE 81 Test for convergence, absolute convergence and conditional convergence of the series,

1 1 11 ..........5 9 13

− + − + (JNTU 2003)

SOLUTION

The given alternating series is of the form 1( 1) ,nnu−−∑ where,

14 3nu

n=

−.

Hence, 0 ;nu n N> ∀ ∈ ( )1

1 14 1 3 4 1nu

n n+ = =+ − +

11 1

4 3 4 1n nu un n+− = −

− +

4 1 4 3

(4 3)(4 1)n nn n

+ − +=

− +

4 0,(4 3)(4 1)

n Nn n

= > ∀ ∈− +

i.e., 1,n nu u n N+> ∀ ∈ 1 0;

4 3nn nLt u Lt

n→∞ →∞= =

−

All conditions of Leibnitz’s test are satisfied.

Hence 1( 1)nnu−−∑ is convergent.

1 ;

4 3nun

=−

Take 1

nvn

= ; 1 03 4(4 )

n

n nn

u nLt Ltv n n

→∞ →∞= = ≠

− and finite.

∴ By comparison test, nu∑ and nv∑ behave alike.


67

But by p − series test, nv∑ is divergent (since 1)p = .

nu∑ is divergent and ∴ The given series is conditionally convergent.

EXAMPLE 82

Test the series 1

1

1( 1) .3

n

n n

∞−

=

−∑ , for absolute / conditional convergence .

SOLUTION The given series is an alternating series of the form ( ) 11 n

nu−−∑ .

Here

(i) 1 ,

3nu n Nn

= ∀ ∈

(ii) ( )3 1 3 3 1 3 ,n n n n n+ > ⇒ + > ∀ .

∴ 1 1

3 1 3n n<

+ , i.e., 1 ,n nu u n N+ < ∀ ∈

And 1 0

3nn nLt u Lt

n→∞ →∞= =

∴ By Leibnitz’s test, the given series is convergent.

But 1 1 1( 1) .3 3

n

n n−− =∑ ∑ is divergent by p − series test (since

p = 1 12

< )

∴ The given series is conditionally convergent.

EXAMPLE 83

Test the following series for absolute / conditional convergence.

(a) 12

1

sin( )( 1) ,n

n

nn

α∞−

=

−∑ (b) 2

13

1( 1) .

1n

n

nn

∞−

=

−+∑

(c) 0

( 1)(2 )!

n

n n

∞

=

−∑ (d) 13 1( 1) .

nn

n

ne

π−+−∑


68

SOLUTION

(a) 2 2

sin 1n

nu

n nα

= < since sin 1nα⎡ < ⎤⎣ ⎦ considering 21

nv n= and using

comparison and p - series tests, we get that nu∑ is convergent nu∑ is absolutely convergent .

(b) By Leibnitz’s test, the series converges.

Taking 1

nvn

= , by comparison and p - series tests , 2

3 1n

n +∑ , is seen to

be divergent. Hence given series is conditionally convergent.

(c) Take 1

2 !nun

= ; 1 0 1n

nn

uLt

u+

→∞= < ; By ratio test, nu∑ is convergent;

Hence given series is absolutely convergent.

(d) 3 1

n

n n

nue

π+= ; By root test, is convergent, ∴ given series is absolutely

convergent. [In problems (a) to (d) above, hints only are given. Students are advised to do the complete problem themselves]

EXAMPLE 84 Find the interval of convergence of the following series.

(a) 13

1( 1)

nn

n

xn

∞−

=

−∑ (b) 1

1

( 2)( 1)3

nn

nn

n x∞−

=

+−∑ (c) log(1 )x+

SOLUTION

(a) Let the given series be nu∑ ; Then 3

n

n

xu

n= ;

1

1 3( 1)

n

n

xu

n

+

+ =+

3

1 .1

n

n nn

u nLt Lt xu n

+

→∞ →∞

⎛ ⎞= ⎜ ⎟+⎝ ⎠ =

3

1 .11nLt x x

n→∞

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟+⎝ ⎠

∴ By ratio test, nu∑ is convergent if 1x <

i.e., nu∑ is absolutely convergent if 1x < ;


69

∴ nu∑ is convergent if 1x <

If x = 1, the given series becomes 3 3 3

1 1 11 ........2 3 4

− + − +

which is convergent, since 3

1n∑ is convergent.

Similarly, if x = –1, the series becomes 3 3

1 1n n

− = −∑ ∑ which is also

convergent.

Hence the interval of convergence of nu∑ is ( )1 1x− ≤ ≤

(b) Proceeding as in (a),

1 23

n

nn

u xLt

u+

→∞

+=

∴ nu∑ is convergent if 2 3x + < , i.e., if 3 2 3x− < + < , i.e., if

5 1x− < < .

If x = –5, ( )2 11 .nnu n−= −∑ ∑ , and is divergent ( in both these cases

If x = 1, ( ) 11 .nnu n−= −∑ ∑ , and is divergent 0nn

Lt u→∞

≠ )

Hence the interval of convergence of the series is (–5 < x < 1)

(c) log (1 + x ) = 2 3 4

.......2 3 4x x xx − + − +

= ( ) 1

11

nn

nn

x un

∞−

=

− =∑ ∑ (say)

n

n

xu

n= ;

1

1 1

n

n

xu

n

+

+ =+


70

1 1 | |11

n

n nn

uLt Lt x x

un

+

→∞ →∞= =

⎛ ⎞+⎜ ⎟⎝ ⎠

By ratio test, nu∑ is convergent when 1x <

i.e., nu∑ is absolutely convergent and hence convergent when –1 < x < 1.

When x = –1, ( ) 1 1 1 1 11 . 1 .......2 3 4

nnu

n−= − = − + − +∑ ∑ ,

which is convergent by Leibnitz’s test. (give the proof )

When x = 1, nu∑ = 1 1 11 ......2 3 4

⎛ ⎞− + + + +⎜ ⎟⎝ ⎠

which is divergent,

since 1n∑ is divergent by p –series test ( prove ) .

Hence nu∑ is convergent when 1 1x− < ≤ .

Interval of convergence is ( 1 1x− < ≤ ).

Exercise 1.5

1. Use integral test and determine the convergence or divergence of the following

series:

1 2

1n∑ ...................................... [Ans : convergent]

2. ( )2

2

1logn n

∞

∑ ....................................... [Ans : convergent]

2. Test for convergence of the following series:

1. 1 1 11 ....2 4 6

− + − + .................................. [Ans : convergent]

2. ( )

( )( )

1

1

12 1 2

n

n n

−∞ −−∑ ................................... [Ans : convergent]


71

3. ( )51 2

11 n n

∞−−−∑ ................................... [Ans : convergent]

3. Classify the following series into absolutely convergent and conditionally

convergent series :

1. ( )

3

1 n

n−

∑ ....................................... [Ans : abs.cgt]

2. 32

sin n

n∑ .......................................... [Ans : abs.cgt]

3. ( )( )2

1log

n

n n−

∑ ......................................... [Ans : abs.cgt]

4. Find the interval of convergence of the following series :

1. 2n nx

n∑ ............................................ [Ans : x−∞ < < ∞ ]

2. 2

nxn∑ ............................................ [Ans : 1 1x− ≤ ≤ ]

3. 2 3 4

......2 3 4

x x xx − + − + .................................. [Ans : 1 1x− < ≤ ]

5. (a) Show that 2 2 2

1 1 11 ....2 3 4

− + − + is absolutely convergent.

(b) Show that 1 1 11 ....2 3 4

− + − + is conditionally convergent.

Summary

1. The geometric series 1

1

n

nx

∞−

=∑ converges if | | 1x < , diverges if 1x ≥ , and oscilates

when 1x ≤ − 2. If nu∑ is convergent, 0nn

Lt u→∞

= [The convergent need not be necessary ]

3. p – series test :- 1

1p

n n

∞

=∑ is convergent if p > 1 and divergent if 1p ≤


72

4. Comparison test :- The series nu∑ and nv∑ are both convergent or both

divergent if n

nn

uLtv→∞

is finite and non – zero.

5. D’Alembert’sRatio test :- nu∑ converges or diverges according as

1 1 or 1n

nn

uLtu

+

→∞< >

1

Alternately, if 1 or 1n

nn

uLtu→∞

+

⎛ ⎞> <⎜ ⎟

⎝ ⎠ . If the limit = 1, the test fails

6. Raabe’s test : nu∑ converges or diverges according as

1

1 1 or 1 n

nn

uLt nu→∞

+

⎡ ⎤⎧ ⎫− > <⎨ ⎬⎢ ⎥

⎩ ⎭⎣ ⎦.

7. Cauchy’s root test: nu∑ converges or diverges according as 1

1 or 1 nnn

Lt u→∞

⎛ ⎞< >⎜ ⎟

⎝ ⎠ (If limit = 1, the test fails.)

8. Integral test : A series ( )nφ∑ of +ve terms where ( )nφ decreases as n increases

is convergent or divergent according as the integral ( )1

x dxφ∞


9. Alternating series – Leibnitz’s test: An alternating series ( ) 1

1

1 nn

n

u∞

−

=

−∑ convergent

if (i) nuu nn ∀> + ,1 and (ii) 0nnLt u→∞

=

10. Absolute / conditional convergence : (a) nu∑ is absolutely convergent if | |nu∑ is convergent.

(b) nu∑ is conditionally convergent if nu∑ is convergent and | |nu∑ is

divergent.

(c) An absolutely convergent series is convergent, but converse need not be true .

i.e., a convergent series need not be convergent.


73

Miscellaneous Exercise - 1.6 1. Examine the convergence of the following series:

1. 1 1 1 ......

1.2 3.4 5.6+ + + ................................... ... [cgt.]

2. 2 2 2

3 3 3

1 2 3 ......1 1 2 1 3 1

+ + ++ + +

.............................. [dgt.]

3. 2 32 2 2 .......

1 2 3+ + + .............................. [dgt.]

4. 2 3

1 2 3 .....1 2 1 2 1 2

+ + ++ + +

................................ [cgt.]

5. ( )2 3

2 3 ..... 01 1 1

x x x xx x x

+ + + >+ + +

........ [cgt. if 1x ≤ dgt. if x > 1]

6. ( )2 33 42 ..... 0

8 27x xx x+ + + > .......................... [cgt. if 1x ≤ dgt. if x > 1]

7. 1 1.3 1.3.51 ....2 2.4 2.4.6

+ + + + ................................. [dgt.]

8. 2 2 2 2 2 2

2 2 2 2 2 2

3 3 .5 3 .5 .7 ......6 6 .8 6 .8 .10

+ + + ............................... [cgt]

9. 3.4 4.5 5.6 ....1.2 2.3 3.4

+ + + ................................ [dig.]

10. ( ) ( ) ( ) ( )

2 2 32 31 2 3

. .... 02 4 6

x x x x+ + + > ............. [cgt. if x <4, dgt. if 4x ≥ ]

11. ( )2 3

2 2 21 ..... 02 3 4x x x x+ + + + > ......................... [cgt.if 1x ≤ , dgt. if x > 1]

12. ( )2 3

2 33 4 5 ..... 04 5 6x x x x⎛ ⎞ ⎛ ⎞+ + + + >⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ............. [cgt if x < 1, dgt. if 1x ≥ ]

13. 2

11n

n⎛ ⎞+⎜ ⎟⎝ ⎠

∑ .......................... [dgt.]


74

14. 3

2

23

n

n∑ ................................................ [cgt.]

15. 2 , 11

na an

<+∑ ................................................ [cgt.]

16. 1 1 11 .....

2.2 3.3 4.4− + − + − + − ....................... [Abs. cgt.]

2. Examine for absolute and conditional convergence of the following series:

1. ( )3

2

313

nn

n−∑ .................................... [Abs. cgt.]

2. ( )1 .

2

n

n

n−∑ ...................................... [Abs. cgt.]

3. 1 1 1 112 3 4 5

− + − + ........................... [Cond. cgt]

4. ( ) ( )2

3

11 n n

n+

−∑ .............................. [Cond.cgt.]

3. Determine the interval of convergence of the following series :

1. 2 3 4

.....2 3 4x x xx + + + + ............................... [ 1 1x− ≤ < ]

2. ( )1

.2

n

n

xn+

∑ ................................ [ 3 1x− < < ]

Solved University Questions (J.N.T.U)

1. Test the convergence of the series:

1 2 3 .........

1.2.3 2.3.4 3.4.5+ + +

Solution Let nu be the thn term of the series;

Then, ( )( ) ( )( )

11 2 1 2nnu

n n n n n= =

+ + + +


75

Let 2

1nv

n= ; then,

( )( )2

1 2n

n nn

u nLt Ltv n n→∞ →∞

=+ +

= 1 1

1 21 1nLt

n n→∞

=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

,

Which is non-zero and finite.

∴ By comparison test, both nu∑ and nv∑ converge or diverge together.

But nv∑ is convergent by p-series test (p > 1) ∴ nu∑ is convergent.

2. Show that every convergent sequence is bounded Solution

Let na be a sequence which converges to a limit ‘l ‘say.

nnLt a l→∞

= ⇒ given any +ve number ∈ , however small ,

we can always find an integer ‘ m ‘ , , na l∋ − <∈ , n m∀ ≥

Taking ∈ = 1, we have, 1na l− < ;

i.e., ( ) ( )1 1nl a l− < < + , n m∀ ≥

Let λ = min ( ){ }1 2 1, ,........ , 1ma a a l− − , and μ = max ( ){ }1 2 1, ,........ , 1ma a a l− +

Then obviously, naλ μ≤ ≤ , n N∀ ∈ ;

Hence na is bounded.

3. Show that the geometric series 2

0

1 .......m

m

q q q∞

=

= + + +∑ converges to the sum

11 q−

when 1q < and diverges when 1q ≥

(JNTU 2001) Solution See theorm 1.2.3 (replace ‘x ‘ by ‘q’ ) .


76

4. Define the convergence of a series. Explain the absolute convergence and conditional convergence of a series. Test the convergence of series

(JNTU 2000)

2

11n

n

−⎡ ⎤

+⎢ ⎥⎣ ⎦∑

Solution For theory part , refer 1.2.1, 1.2.2, 1.8, 1.8.1, 1.9.1 and 1.9.2

Problem : Let

2

11n

nun

−⎛ ⎞= +⎜ ⎟⎝ ⎠

; ( )1 11n

nnn n

Lt u Ltn

−

→∞ →∞

⎛ ⎞= +⎜ ⎟⎝ ⎠

= 2

1 1 111

nnLt

e

n

→∞= <

⎡ ⎤+⎢ ⎥⎣ ⎦

By Cauchy’s root test, nu∑ is convergent.

5. Test the convergence of the series, 2 31 1.3 1.3.51 .....2 2.4 2.4.6

x x x+ + + +

(JNTU 2001) Given that x > 0. Solution Omitting the first term of the series, we have,

( )1.3.5. 2 1

2.4.6....2n

n

nu x

n−

= ; ( )( )

11

1.3.5..... 2 1.

2.4.6..... 2 2n

n

nu x

n+

+

+=

+;

1 2 1 .2 2

n

n nn

u nLt Lt x xu n

+

→∞ →∞

+⎛ ⎞= =⎜ ⎟+⎝ ⎠

By ratio test, nu∑ is convergent when x < 1, and divergent when x > 1

The ratio test fails when x = 1

When x = 1, 1

2 2 11 12 1 2 1

n

n

u nu n n+

+− = − =

+ +

1

11 12 1 2

n

n nn

u nLt n Ltu n→∞ →∞

+

⎡ ⎤⎛ ⎞ ⎛ ⎞− = = <⎢ ⎥⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠⎣ ⎦ ;

∴ By Raabe’s test , nu∑ diverges . ∴ The given series converges when x < 1 and diverges when 1x ≥ .


77

6. Test the convergence of the series, 2 3

2 31 2 3 4 ......... 02 3 4 5

x x x x⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + >⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(JNTU 2002) Solution Neglecting the 1st term,

12

n

nnu xn

⎡ + ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦;

1

11122 1

nn

n nu x xn n

⎛ ⎞++⎛ ⎞ ⎜ ⎟= =⎜ ⎟+ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

1n

nnLt u x→∞

= ; By Cauchy’s root test, nu∑ is cgt. when x < 1 and dgt. when

x > 1; when x = 1 , the test fails .

When x = 1, ( )( )

11

21

n

n nnu

n

+=

+; 2

1 0nn

eLt ue e→∞

= = ≠

∴ nu∑ is divergent.

∴ ∑ nu is cgt. when x < 1 and dgt. when 1x ≥ .

7. Test the series whose thn term is ( )3 1 2nn − for convergence. (JNTU 2003) Solution

( )3 1

2n n

nu

−= ;

( ){ }1 1

3 1 12n n

nu + +

+ −= ;

( )( )

1 3 22 3 1

n

n

nuu n

+ +=

− 1 1 1

2n

nn

uLtu

+

→∞= < ;


8. Show by Cauchy’s integral test that the series ( )2

1log p

n n n

∞

=∑ converges if p > 1

and diverges if 0 1p< ≤ (JNTU 2003) Solution

Let ( )( )

1log px

x xφ = ; 2x ≥ ; Then ( )xφ decreases as x increases in [ ]2,∞


78

( )( )2 2 log p

dxx dxx x

φ∞ ∞

=∫ ∫ log 2

p

duu

∞

= ∫ 1

log 21

pup

−∞=

− ;

[Taking log x = u, 1 dx dux

= 2 log 2x u= ⇒ = and x u= ∞ ⇒ = ∞ ]

Case (i) : 1 1 0p p> ⇒ − < ⇒ Integral is finite , and Case (ii) : 0 1p< ≤ ⇒ Integral is infinite.

Hence, by integral test, the given series converges if p > 1 and diverges when 0 1p< ≤ .

9. Test the convergence of the series ∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛+

23

11n

n

Solution

;11

111

12

3

1

n

nn

nn

n

nu

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛+

1 1 1nnn

Lt ue→∞

= < [ 2 < e < 3 ] .


10. Test the convergence of the series,( )

( )2

1 .1

n n

n

xn n

∞

=

−−∑ , 0 < x < 1

Solution

The given series is of the form ( )1 nnu−∑ , where

( )1

n

nxu

n n=

− .

This is an alternating series in which (i) 0nu > and 1n nu u n N+> ∀ ∈ . Further 0nn

Lt u→∞

= . Hence, by Leibnitz test, the series is convergent.

11. Discuss the convergence of the series, 2 4 61 ..........

2 1 3 2 4 3 5 4x x x

+ + + +

(JNTU 1995, 2002, 2003, 2008) Solution

thn term of the series = ( )

2

2 1

n

nxu

n n=

+ + (omitting 1st term)


79

( )

2 2

1 3 2

n

nxu

n n

+

+ =+ +

; ( )

21 2 1 .3

n

n

u n n xu n

+ + +=

+

( )

2 212 11 . 1

.31

n

n nn

u n nLt Lt x xu

n

+

→∞ →∞

⎡ ⎤+ +⎢ ⎥= =⎢ ⎥+⎢ ⎥⎣ ⎦

;

∴ By ratio test, nu∑ converges if 2 1x < , i.e., if 1x < , and diverges if

2 1x > , i.e., if 1x > ;

When 2 1x = , ( )

12 1nu

n n=

+ +; taking 3

2

1nv

n= ,

( )

32

32

12 11 1

n

n nn

u nLt Ltv n n n

→∞ →∞= =

+ +

∴ By comparison test, nu∑ and nv∑ both converge or diverge together;

But nv∑ is convergent by p-series test.

∴ nu∑ is convergent if 1x ≤ and divergent if 1x > .

12. Test the convergence of the series ( )

2

1 1

n

n

xn n

∞

= +∑ (JNTU 2006, 2007)

Solution

( )

2

1

n

nxu

n n=

+;

( )

2 2

1 2 1

n

nxu

n n

+

+ =+ +

( )

2 21111 . .

22 1n

n

u n n nx xu n

n

+++

= =+ +

; 21n

nn

uLt xu

+

→∞= ;

∴ By ratio test, nu∑ converges when 1x < and diverges for 1x > .

When 1x = , ( )3

2

111

nun n

=+

taking nv 32

1

n= and applying the comparision

test, we observe that nu∑ is convergent .

Hence nu∑ converges when 1x ≤ and diverges when 1x > .


80

13. Find the interval of convergence of the series, 2 3 4

.........2 3 4x x x

+ + + ∞

(JNTU 2006, 2007)

Solution

For the given series, 1

1

n

nxun

+

=+

; 2

1 2

n

nxun

+

+ =+

11121

n

n nn

u nLt Lt x xu n

+

→∞ →∞

⎛ ⎞+⎜ ⎟= =⎜ ⎟+⎝ ⎠

By ratio test, nu∑ converges when 1x < i.e., 1 1x− < <

When x = 1, 1

1nun

=+

Taking 1

nun

= ; 111

n

n

uv n

=+

and, 1 0n

nn

uLtv→∞

= ≠ and finite.

∴ Both nu∑ and nv∑ converge or diverge together.

But nv∑ diverges ∴ nu∑ also diverges when x = 1.

When x = –1, the given series is

1 1 1 12 3 4 5

− + − + which is alternating series with

1n nu u n+> ∀ and 0nu → as n → ∞

∴ By Leibneitz’s test nu∑ converges when x = –1

∴ Interval of convergence is [(–1, 1) i.e., 1 1x− ≤ <

14. Test the convergence of the series ( )( )1

1.3.5.... 2 12.5.8.... 3 2n

nn

∞

=

++∑ (JNTU 2006)

Solution

( )( )

1.3.5.... 2 12.5.8.... 3 2n

nu

n+

=+

; ( )( )1

1.3.5.... 2 32.5.8.... 3 5n

nu

n+

+=

+


81

1 2 33 5

n

n

u nu n

+ +=

+;

( )( )

132 2 15 33

n

n nn

u nLt Ltu

n

+

→∞ →∞

⎡ ⎤+⎢ ⎥= = <⎢ ⎥+⎣ ⎦


15. Prove that the series ( )( )∑

∞

=22

1

n

n

nlogn– converges absolutely. (JNTU 2006)

Solution

( )3

1log

nun n

= ; ( )3 2

2 log 2logdx dt

tx x

∞ ∞

=∫ ∫

(where t = log x ) log 21 1

log 2t∞−

= = , which is finite.

∴ By integral test nu∑ is convergent.

∴ nu∑ converges absolutely.

Note: This problem can also be done by Leibnitz test. The reader is advised to try that method also.


3

2 1, 0

1nn

x xn

+>

+∑ (JNTU 2006)

Solution

thn term of the given series , 3

2 11

nn

nu xn

+=

+;

( )

( )1

1 3

2 1 11 1

nn

nu x

n+

+

⎡ ⎤+ += ⎢ ⎥

+ +⎢ ⎥⎣ ⎦

( )1

3

2 31 1

nn xn

++=

+ +

( )( ){ }

( )( )

1 31

3

12 3 .2 11 1

n

nnn n

n

nn xuLt Ltu x nn

+

+

→∞ →∞

++= ×

++ +

( ) ( )

( ) ( )

33

33

3

3 12 1 . 12 .11 11 .2 1 2

n

n nn nLt x xn nn nn

→∞

⎡ ⎤+ +⎢ ⎥

⎢ ⎥ =⎧ ⎫⎢ ⎥+ + +⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦

By ratio test, nu∑ converges if x < 1 and diverges if x > 1. If x = 1 the test fails.


82

When x = 1, 3

2 11n

nun

+=

+; Taking 2

1nv

n= ;

23

2 1 2 01

n

n nn

u nLt Lt nv n→∞ →∞

+= × = ≠

+ and finite

∴ nu∑ and nv∑ converge or diverge together .

But nv∑ converges ∴ nu∑ also converges.

Thus, nu∑ converges when 1x ≤ and diverges when x > 1.

17. Test the series ( ) ( )

21

1 logn

n

nn

∞

=

−∑ , for absolute/conditional convergence

(JNTU 2006) Solution

( ) ( )

2

1 logn

n

nu

n−

= ; ( )

2

logn

nu

n= ;

22 log 2

log tx dx te dtx

∞ ∞−=∫ ∫ [taking log x = t , tx e= , 1 logdx tx = ]

[ ] ( )log 2log 2

10 1 log 2 . log 2 12

t tte e e− − ∞ −= − + = − − = − ,

which is finite.

∴ By integral test nu∑ is convergent nu⇒ ∑ converges absolutely.

(Note that nu∑ is cgt. by Leibnitz’s test).


1log log nn

∑ (JNTU 2006)

Solution

Given that ( )

1log log

n nun

= ;

1 1 0 1

log logn

nn nLt u Lt

n→∞ →∞

⎡ ⎤= = <⎢ ⎥

⎣ ⎦



83

19. Find the interval of convergence of the series,

3 5 71 1 3 1.3.5. . . . ......

2 3 2 4 5 2.4.6 7x x xx + + + + (JNTU 2007)

Solution

Term of the series, ( )

( )2 11.3.5..... 2 1

.2.4.6.....2 2 1

n

n

n xun n

+−=

+ (neglecting 1st term)

( )( )

( ) ( )2 3

1

1.3.5..... 2 1 2 1.

2.4.6.....2 2 2 2 3

n

n

n n xun n n

+

+

− +=

+ + ;

( )

( )( )

221 2 1

.2 2 2 3

n

n nn

nuLt Lt xu n n

+

→∞ →∞

⎡ ⎤+= ⎢ ⎥

+ +⎢ ⎥⎣ ⎦

( )( )

22

2 2

22

4 14.

10 64n

n n nLt x xn n n

→∞

⎡ ⎤+ +⎢ ⎥ =⎢ ⎥+ +⎢ ⎥⎣ ⎦

By ratio test, nu∑ converges when 2 1x < , i.e., 1 1 1x x< ⇒ − < <

When 2 1x = , the test fails;

Then 2

2 21

4 10 6 8 51 14 2 1 4 2 1

n

n

u n n nu n n n n+

⎛ ⎞+ + +− = − =⎜ ⎟+ + + +⎝ ⎠

( )

( )2

21 2

581 2 1

2 14n

n nn

nu nLt n Ltu n n n

→∞ →∞+

+⎡ ⎤⎛ ⎞− = = >⎢ ⎥⎜ ⎟

⎝ ⎠ + +⎣ ⎦

∴ By Raabe’s test, nu∑ converges when 2 1x = , i.e., 1x = ± .

∴ Interval of convergence of nu∑ is [–1 < x < 1]

20. Test the series 2

1 2 3 ... ,2 3 8 –1

nn

+ + + + for convergence or divergence.

[JNTU 2006] Solution

un = 2 1n

n + ; Let νn = 3/2

1n

n

n

uν

= 3/2

2

. 1

n nn +

=2

2 1n

n + =

2

111n

+


84

2

1 = 111+

n

n nn

uLt Lt

nν→∞ →∞

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

which is non-zero finite number

∴ By comparison test, Σun and Σνn behave alike.

But Σνn is convergent by p-series test 3p 12

⎛ ⎞= >⎜ ⎟⎝ ⎠∵

Hence Σun is convergent

21. Test the convergence of the series, 2 2 2

2 1 3 1 4 1 .......3 1 4 1 5 1

− − −+ + +

− − −

[JNTU 2007] Solution

un = 2

1 1 ;( 2) –1

n+n

−+

Let νn = 3/2

1n

[∵ Highest degree of n in Dr – Nr = 2 – 1/2 = 3/2]

n

n

uν

= ( )3/2

2

. 1 1

2) –1

n n+

(n

−

+ =

2

22

1 11

4 31

nn n

nn n

= 1 nnn

n

uLt uν→∞

⇒ Σ and Σνn both converge or diverge together (by comparison

test). But Σνn converges by p-series test 3p 1 .2

⎛ ⎞= >⎜ ⎟⎝ ⎠∵ Hence Σun is

convergent.

22. Test the convergence of 3 3 1n nΣ + − [JNTU 2008]

Solution

un can be written as, 3 3 3 3

3 3

1 1

1n

n n n nu

n n

i.e. un = 3

3

111+ +1nn

⎡ ⎤⎢ ⎥⎣ ⎦

; Let νn = 3/2

1n


85

Then, n

n

uν

=

3

111+ +1n

⇒ 1 0.2

n

nn

uLtν→∞

= ≠

∴ By comparison test, Σun and Σνn have same property

Σνn is convergent by p-series test 3p 1 .2

⎛ ⎞= >⎜ ⎟⎝ ⎠∵ Hence Σun is convergent.

23. Test for the convergence of the series, 2 3

1 .... 02 5 10x x x x+ + + + >

[JNTU 1998, 1985, 2002, 2002] Solution

Neglecting the first term, we observe that the nth term of the series,

un = 2 ;1

nxn +

un+1 = 1

2 ,2 2

nxn n

+

+ + so that,

2

12

+1 = + 2 + 2

n

n nn

u nLt Lt x xu n n

+

→∞ →∞

⎛ ⎞= ⎜ ⎟

⎝ ⎠

∴ By ratio test, Σun converges when x < 1 and diverges when x > 1 and the test fails when x = 1

∴ when x = 1, un = 2

11n +

; Taking Σνn = 2

1n

and using comparison test, we can

show that Σun is convergent. [This part of proof is left to the reader as an exercise].

∴ Σun converges for x < 1 and diverges for x > 1.

24. Test the convergence of 2

. 1

nn xn

Σ+

(x > 0) [JNTU 2003]

Solution

2

12

+1 1 . 2 2

n

n nn

u n nLt Lt . xu nn n

+

→∞ →∞

+=

+ +

= 2

2

11+ . 1 1 . 1. 2 21

n

xnLt x x

nn n

→∞+ = =

+ +


86

∴ By ratio test, Σun converges when x < 1 and diverges when x > 1 and when x = 1 the test fails.

When x = 1, un = 2 1n

n +; Taking νn = 1 ,

n 1 n

nn

uLtν→∞

= (verify)

By comparison test, Σun diverges. [Since Σνn diverges by p-series test as p = 1 12

< ]

Hence Σun converges when x < 1 and diverges when x > 1.

25. Test for convergence the series nx

nΣ (x > 0) [JNTU 2007, 2008]

Solution

un = ,nx

n un+1 =

1

1

nxn

+

+ ; 1

1n

n nn

u nLt Lt x xu n

+

→∞ →∞

⎛ ⎞= =⎜ ⎟+⎝ ⎠

∴ By comparison test, Σun converges when x < 1 and diverges when x > 1. when x = 1, the test fails.

When x = 1, un = 1n

and Σun is divergent (p series test – p = 1)

∴ Σun is convergent when x < 1 and divergent when x > 1.

26. Find whether the series 2

1sin n (–1)

(n–1)n

n

∞

=

⎛ ⎞⎜ ⎟⎝ ⎠Σ is absolutely convergent or

conditionally convergent. [JNTU 2006] Solution When n > 2, we have,

nu =

1sin n

(n–1)

⎛ ⎞⎜ ⎟⎝ ⎠ ; Let νn = 3/2

1n

;

∴ n

n

uν

= 3/2

n–1n 1sin

n⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦


87

1sin = 11 1

n

n nn

u nnLt Ltn

nν→∞ →∞

⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ =⎜ ⎟ ⎜ ⎟−⎝ ⎠⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

∴ By comparison test Σ nu and Σνn behave alike. But Σνn is convergent by

p-series test (p = 3/2 > 1).

∴ Σ nu is convergent. Hence Σun is absolutely convergent.

27. Test whether the series 21

cos n +1n

n π∞

=Σ converges absolutely [JNTU 2006]

Solution

The given series is 21

cos n +1n

n π∞

=Σ =

1( 1) n

nu∞

Σ − where un = 2

1 .1n +

It is obvious that u1 > u2 > …. un > un + 1 > …. and 0nnLt u→∞

=

∴ By Leibnitz’s test given series is convergent.

2

1( 1) . 1

nnu

nΣ − = Σ

+ which is convergent (Take ν = 2

1n

and apply comparison

test. This is an exercise to the reader)

Hence given series is absolutely convergent.

28. Find whether the following series converges absolutely or conditionally

1 1 1 1.3.5 1.3.5.7 . .....6 6 3 6.8.10 6.8.10.12

− + − + [JNTU 2007]

Solution The given series is 16

1 3.5 3.5.71 .....3 8.10 8.10.12

⎡ ⎤− + − +⎢ ⎥⎣ ⎦

= 16

2 3.5 3.5.7 .....3 8.10 8.10.12

⎡ ⎤⎧ ⎫+ − +⎨ ⎬⎢ ⎥⎩ ⎭⎣ ⎦

We can take the series as 3.5 3.5.7 .....,8.10 8.10.12

− + (neglecting other terms) = Σun,

where 3.5.7.....(2 3) ,8.10.12.....(2 8)n

nun+

=+

so that 13.5.7.....(2 3) (2n 5)

8.10.12.....(2 8) (2n 10)nnun+

+ +=

+ +


88

It can be seen that 1

n

nn

uLtu→∞

+

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

= (2 10) 1,(2 5)n

nLtn→∞

⎡ ⎤+=⎢ ⎥+⎣ ⎦

so that ratio test fails.

1

1n

nn

uLt nu→∞

+

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠ = (2 10) 1

(2 5)n

nLt nn→∞

⎡ ⎤+−⎢ ⎥+⎣ ⎦

= 5 5 12 5 2n

nLtn→∞

⎛ ⎞ = >⎜ ⎟+⎝ ⎠

∴ By Raabe’s test, nuΣ is convergent. Hence the given series is absolutely

convergent.

29. Test for absolute convergence of the series whose nth term is ( 1) (x 2) .2 5

n

n

− ++

[JNTU 2007]

Solution: Let given series be Σun. Then 22 5n n

xu +=

+

∴ 1 1

2 = ,2 5n n

xu + +

++

= so that, 11

52 12 5 2 = =

52 5 2 22

nn n

nn

nnn

uu

++

⎛ ⎞+⎜ ⎟+ ⎝ ⎠+ ⎛ ⎞+⎜ ⎟

⎝ ⎠

∴ 1 1 1 ;2

n

nn

uLtu

+

→∞= <

∴ By ratio test, nuΣ is convergent. Hence the given series is absolutely

convergent.

30. Test whether the following is absolutely convergent or conditionally

convergent.

( )1

1 ( 1) n 1n n

∞+Σ − + − [JNTU 2008]

Solution

The given series is Σun where un = (– 1)n+1 n 1 .n⎡ ⎤+ −⎣ ⎦ It is an alternating

series.

n 1 n+ − = n 1 n 1 1

n 1 n 1 n

n n

n nν

⎡ ⎤ ⎡ ⎤+ − + +⎣ ⎦ ⎣ ⎦ = =+ + + +

(say),


89

(1) νn > 0 ∀ n ; (2)nLt→∞

νn = 0 and (3) νn + 1 > νn ∀ n; since all conditions of

Leibnitz’s test are satisfied, the given series is convergent.

Further, 1 = ;n 1nu

n+ + Taking 1 ,

nnν = we have,

1 = 0211 1

n

n nn

u nLt Ltn

nν→∞ →∞

= ≠⎡ ⎤

+ +⎢ ⎥⎣ ⎦

∴ By comparison test, nuΣ and Σνn behave alike. But Σνn = 1 n

Σ is

divergent by p-series test 1 1 .2

p⎛ ⎞= <⎜ ⎟⎝ ⎠

∴ nuΣ is divergent. Hence the given series is conditionally convergent

31. Find the interval of convergence of the series,

2 3

1 1 1 ........1 2(1 ) 3(1 )x x x

+ + +− − −

[JNTU 2007, 2008]

Solution: If the nth term of the given series is un,

un = 1(1 )nn x−

; un+1 = 1

1( 1) (1 )nn x ++ −

1+1

1 (1 ) = ( +1) (1 – ) 1

nn

nn nn

u n xLt Ltu n x

+

→∞ →∞

⎡ ⎤ −⎢ ⎥⎣ ⎦

= 1 1 1 . = 1 (1 ) 11

nLt

x xn

→∞ − −⎛ ⎞+⎜ ⎟⎝ ⎠

Now, by ratio test:

(i) x = 1 ⇒ the limit is infinite ⇒ Σun is divergent

(ii) x ≠ 0 and x > 1 ⇒ the limit is < 1 ⇒ Σun is convergent


90

(iii) x ≠ 0 and x < 1 and > 0 ⇒ the limit is > 1 ⇒ Σun is divergent

(iv) If x = 0, the series is 1 + 1 1 .....,2 3

+ + which is divergent by p-series test

(p = 1)

(v) If x < 0, the limit is < 1 ⇒ Σun is convergent by ratio test .

Hence the given series converges for all values of (i) x > 1 and x ≠ 0 and (ii) x < 0.

∴ The interval of convergence of the given series is (– ∞ , 0) ∪ (1, ∞ ).

Objective Type Questions

1. The infinite series 21

11n n

∞

= +∑ is

(i) convergent (ii) divergent (iii) oscillatory (iv) none of these [Ans :(i)]

2. The series 2

11

nn

++

is

(i) convergent (ii) divergent

(iii) oscillatory (iv) none of these [Ans :(ii)]

3. The series 1 1 1 1 .....

1 1 2 2 3 3 4 4− + − +

(i) oscillatory (ii) absolutely convergent

(iii) conditionally convergent (iv) none of these [Ans :(ii)]

4. The series 1 1 11 ......2 3 4

− + − is

(i) oscillatory (ii) divergent

(iii) convergent (iv) none of these [Ans :(iii)]


91

5. The interval of convergence of the series 2 3 4

......2 3 4x x xx − + − + ,is

(i) x−∞ < < ∞ (ii) 1 2x− < <

(iii) 1 1x− < ≤ (iv) none of these [Ans :(iii)]

6. The series 1 2 3 .........

1.2 3.4 5.6+ + + ∞ is


(iii) oscillatory (iv) none of these [Ans :(ii)]

7. The series 1 2 3 .........

1.3 3.5 5.7+ + + ∞ , is

(i) conditionally convergent (ii) convergent

(iii) divergent (iv) none of these [Ans :(iii)]

8. The series 2 3 4 5 ........

1 2 3 4p p p p+ + + + ∞ is convergent if

(i) p < 2 (ii) p = 2

(iii) p > 2 (iv) none of these [Ans :(iii)]

9. The series 6 – 10 + 4 + 6 – 10 + 4 + 6 – 10 + 4 + 6............. ∞ is

(i) convergent (ii) oscillatory

(iii) divergent (iv) none of these [Ans :(ii)]

10. The series 1 1 1 .........

2.4 4.6 6.8+ + + is


(iii) oscillatory (iv) none of these [Ans :(i)]


92

2. Indicate whether the following statements are true or false:

1. The series 1

1 2 n−+∑ is convergent . ........................ [False]

2. The series 2

2

52 7nn

++∑ is not convergent . ...................... [True]

3. The series 1 1 1 ........1 2 3

+ + + is divergent.................... [False]

4. The series 3 5

.......,3 5x xx − + − + − converges when 1 1x− ≤ ≤ .. [True]

5. The series ( ) 11

.5

n

nn

−−∑ is absolutely convergent. ........................ [True]

6. The series 2 3 42 3 4 ....x x x x+ + + + ∞ is convergent if x > 1........ [False]

7. The series 2 3

........2 3x xx + + + ∞ is divergent if 1x ≥ ............... [True]

8. The series 3 3 42 3 4

2! 3! 4!1 ......2 3 4 5x x x x+ + + + + + ∞ is convergent

if x > e ........................... [True]

9. The series

32

11n

n

−⎛ ⎞+⎜ ⎟⎝ ⎠

∑ is divergent ............................. [False]

10. The series 2 3

2 31 2 3 4 .......2 3 4 5

x x x⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

is convergent

if x < 1 . ............................. [True]

11. The series ( )2 31 2 3 4 ..... 1x x x x− + − + ∞ < is divergent............ [False]


93

12. The series 2 3 4

2 3 4 ......1 1 1 1

x x x xx x x x

− + − + ∞+ + + +

is convergent ...... [True]

13. The series 3 3 3

sin sin 2 sin 3 ......1 2 3

x x x− + + ∞ converges absolutely...... [True]

14. The series ( ) 11 n

n

−−∑ is conditionally convergent. ........................ [True]

15. The series whose thn term is ( )

23 52 a

nn

+

+ is convergent. [False]

3. Fill in the Blanks:

1. The geometric series 1

1

n

nar

∞−

=∑ converges if___________ . [Ans: | | 1r < ]

2. If a series of +ve terms nu∑ is convergent, nnLt u→∞

= ______. . [Ans: 0]

3. { }3 3

11

nn n

∞

=

+ −∑ is __________. [Ans: convergent.]

4. If ( )

3

1

3 45 p

n

nn

∞

=

−

+∑ is divergent ,value of p is _________. [Ans: 4≤ ]

5. The interval of convergence of nu∑ where 22

2

22

n

nnu xn

⎛ ⎞−= ⎜ ⎟+⎝ ⎠

, is _______.

[Ans: 1 1x− < < ]

6. nu∑ is convergent series of +ve series. Then ( )1 nnn

Lt u→∞

is_________.

[Ans: 1< ]


94

7. The series 8 – 12 + 4 + 8 – 12 – 4 + .......is _________. [Ans : Oscillatory ]

8. If 0,nu n> ∀ and nu∑ is convergent, then 1

1n

nn

uLt nu→∞

+

⎡ ⎤⎧ ⎫−⎨ ⎬⎢ ⎥

⎩ ⎭⎣ ⎦ is _______.

[Ans: 1> ]

9. If the series ( ) ( )1

1 , 0nn n

n

a a n∞

=

− > ∀∑ is convergent, then for all values of n , 1

n

n

aa +

is______. [Ans: 1> ]

10. If 2

111 ,n

nn nn

u Lt un

−

→∞

⎛ ⎞= + =⎜ ⎟⎝ ⎠

_____________. [Ans: 1 e ]

Documents

Chapter 1 Sequences and Series - BS Publications1.3 Tests for the Convergence of an Infinite Series In order to study the nature of any given infinite series of +ve terms regarding